# Calculator search results

Formula
Number of solution
Relationship between roots and coefficients
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 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 3 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \end{array}$
$\begin{array} {l} x = - \dfrac { 10 } { 3 } \\ x = - \dfrac { 7 } { 2 } \end{array}$
$\color{#FF6800}{ 6 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 0$
 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 
$\alpha + \beta = - \dfrac { 41 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { 35 } { 3 } }$
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