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Formula
Solve the quadratic equation
Answer
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Number of solution
Answer
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Relationship between roots and coefficients
Answer
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Graph
$y = 6 \left ( x + 3 \right ) ^ { 2 } + 5 \left ( x + 3 \right ) + 1$
$y = 0$
$x$Intercept
$\left ( - \dfrac { 10 } { 3 } , 0 \right )$, $\left ( - \dfrac { 7 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 70 \right )$
Minimum
$\left ( - \dfrac { 41 } { 12 } , - \dfrac { 1 } { 24 } \right )$
Standard form
$y = 6 \left ( x + \dfrac { 41 } { 12 } \right ) ^ { 2 } - \dfrac { 1 } { 24 }$
$6 \left( x+3 \right) ^{ 2 } +5 \left( x+3 \right) +1 = 0$
$\begin{array} {l} x = - \dfrac { 7 } { 2 } \\ x = - \dfrac { 10 } { 3 } \end{array}$
Find solution by method of factorization
$6 \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } + 5 \left ( x + 3 \right ) + 1 = 0$
$ $ Expand the binomial expression $ $
$6 \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \right ) + 5 \left ( x + 3 \right ) + 1 = 0$
$\color{#FF6800}{ 6 } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \right ) + 5 \left ( x + 3 \right ) + 1 = 0$
$ $ Multiply each term in parentheses by $ 6$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ x } + \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 9 } + 5 \left ( x + 3 \right ) + 1 = 0$
$6 x ^ { 2 } + \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ x } + 6 \times 9 + 5 \left ( x + 3 \right ) + 1 = 0$
$ $ Simplify the expression $ $
$6 x ^ { 2 } + \color{#FF6800}{ 36 } \color{#FF6800}{ x } + 6 \times 9 + 5 \left ( x + 3 \right ) + 1 = 0$
$6 x ^ { 2 } + 36 x + \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 9 } + 5 \left ( x + 3 \right ) + 1 = 0$
$ $ Multiply $ 6 $ and $ 9$
$6 x ^ { 2 } + 36 x + \color{#FF6800}{ 54 } + 5 \left ( x + 3 \right ) + 1 = 0$
$6 x ^ { 2 } + 36 x + 54 + \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) + 1 = 0$
$ $ Multiply each term in parentheses by $ 5$
$6 x ^ { 2 } + 36 x + 54 + \color{#FF6800}{ 5 } \color{#FF6800}{ x } + \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } + 1 = 0$
$6 x ^ { 2 } + 36 x + 54 + 5 x + \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } + 1 = 0$
$ $ Multiply $ 5 $ and $ 3$
$6 x ^ { 2 } + 36 x + 54 + 5 x + \color{#FF6800}{ 15 } + 1 = 0$
$6 x ^ { 2 } + \color{#FF6800}{ 36 } \color{#FF6800}{ x } + 54 \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ x } + 15 + 1 = 0$
$ $ Calculate between similar terms $ $
$6 x ^ { 2 } + \color{#FF6800}{ 41 } \color{#FF6800}{ x } + 54 + 15 + 1 = 0$
$6 x ^ { 2 } + 41 x + \color{#FF6800}{ 54 } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 0$
$ $ Find the sum $ $
$6 x ^ { 2 } + 41 x + \color{#FF6800}{ 70 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \right ) = 0$
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 3 } } \end{array}$
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = - \dfrac { 7 } { 2 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 6 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 6 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 35 } { 3 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 6 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 35 } { 3 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 35 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + \dfrac { 41 } { 12 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 35 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x + \dfrac { 41 } { 12 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 35 } { 3 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x + \dfrac { 41 } { 12 } \right ) ^ { 2 } = - \dfrac { 35 } { 3 } + \left ( \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x + \dfrac { 41 } { 12 } \right ) ^ { 2 } = - \dfrac { 35 } { 3 } + \dfrac { \color{#FF6800}{ 41 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 12 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 35 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 ^ { 2 } } { 12 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 1 } { 144 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 1 } { 144 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 12 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 1 } { 144 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 41 } { 12 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 1 } { 144 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 1 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 12 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 1 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 12 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 12 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 12 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \end{array}$
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = - \dfrac { 7 } { 2 } \end{array}$
Calculate using the quadratic formula
$\color{#FF6800}{ 6 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = 0$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 41 \pm \sqrt{ 41 ^ { 2 } - 4 \times 6 \times 70 } } { 2 \times 6 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 41 \pm \sqrt{ 1 } } { 2 \times 6 } }$
$x = \dfrac { - 41 \pm \sqrt{ \color{#FF6800}{ 1 } } } { 2 \times 6 }$
$n $ square root of 1 is 1 $ $
$x = \dfrac { - 41 \pm \color{#FF6800}{ 1 } } { 2 \times 6 }$
$x = \dfrac { - 41 \pm 1 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } }$
$ $ Multiply $ 2 $ and $ 6$
$x = \dfrac { - 41 \pm 1 } { \color{#FF6800}{ 12 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 41 \pm 1 } { 12 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 41 + 1 } { 12 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 41 - 1 } { 12 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 41 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } { 12 } \\ x = \dfrac { - 41 - 1 } { 12 } \end{array}$
$ $ Add $ - 41 $ and $ 1$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 40 } } { 12 } \\ x = \dfrac { - 41 - 1 } { 12 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 40 } { 12 } } \\ x = \dfrac { - 41 - 1 } { 12 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { - 10 } { 3 } } \\ x = \dfrac { - 41 - 1 } { 12 } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 3 } \\ x = \dfrac { - 41 - 1 } { 12 } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 3 } } \\ x = \dfrac { - 41 - 1 } { 12 } \end{array}$
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 41 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 12 } \end{array}$
$ $ Find the sum of the negative numbers $ $
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 42 } } { 12 } \end{array}$
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = \color{#FF6800}{ \dfrac { - 42 } { 12 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = \color{#FF6800}{ \dfrac { - 7 } { 2 } } \end{array}$
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 7 } } { 2 } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 6 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 41 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 70 }$
$D = \color{#FF6800}{ 41 } ^ { \color{#FF6800}{ 2 } } - 4 \times 6 \times 70$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 1681 } - 4 \times 6 \times 70$
$D = 1681 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 70 }$
$ $ Multiply the numbers $ $
$D = 1681 \color{#FF6800}{ - } \color{#FF6800}{ 1680 }$
$D = \color{#FF6800}{ 1681 } \color{#FF6800}{ - } \color{#FF6800}{ 1680 }$
$ $ Subtract $ 1680 $ from $ 1681$
$D = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = - \dfrac { 41 } { 6 } , \alpha \beta = \dfrac { 35 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 6 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = 0$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 41 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 70 } = \color{#FF6800}{ 0 }$
$ $ In the quadratic equation $ ax^{2}+bx+c=0 $ , if the two roots are $ \alpha, \beta $ , then it is $ \alpha + \beta =-\dfrac{b}{a} $ , $ \alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 41 } { 6 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 70 } { 6 } }$
$\alpha + \beta = - \dfrac { 41 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { 70 } { 6 } }$
$ $ Reduce the fraction $ $
$\alpha + \beta = - \dfrac { 41 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { 35 } { 3 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-If the sum of two consecutive 
numbers is $45$ and one number is $X$ 
.This statement in the form of 
equation $1s:$ 
$\left(1$ Point) $\right)$ 
$○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ 
$○sx+1ef\left(x+2$ $r1gnt\right)=145s$ 
$sx+1x=45s$
7th-9th grade
Algebra
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ 
$s1S$ 
$S-1S$ 
$s2S$ 
$s50s$
7th-9th grade
Other
search-thumbnail-Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ 
Find f(7)fAleft(7\right) 
O a 
O b -8 
6. 
$5$
7th-9th grade
Algebra
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