Symbol

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Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = - x ^ { 2 } + 7 x - 10$
$y = 0$
$x$Intercept
$\left ( 2 , 0 \right )$, $\left ( 5 , 0 \right )$
$y$Intercept
$\left ( 0 , - 10 \right )$
$-x ^{ 2 } +7x-10 = 0$
$\begin{array} {l} x = 5 \\ x = 2 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 2 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = - 10 + \left ( \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
 When raising a fraction to the power, raise the numerator and denominator each to the power 
$\left ( x - \dfrac { 7 } { 2 } \right ) ^ { 2 } = - 10 + \dfrac { \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 ^ { 2 } } { 2 ^ { 2 } } }$
 Organize the expression 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 9 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 9 } { 4 } }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 9 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 9 } { 4 } } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \end{array}$
 Organize the expression 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = 2 \end{array}$
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 7 \right ) \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 10 } } { 2 \times 1 }$
 Simplify Minus 
$x = \dfrac { 7 \pm \sqrt{ \left ( - 7 \right ) ^ { 2 } - 4 \times 1 \times 10 } } { 2 \times 1 }$
$x = \dfrac { 7 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 10 } } { 2 \times 1 }$
 Remove negative signs because negative numbers raised to even powers are positive 
$x = \dfrac { 7 \pm \sqrt{ 7 ^ { 2 } - 4 \times 1 \times 10 } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 \pm \sqrt{ 7 ^ { 2 } - 4 \times 1 \times 10 } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 \pm \sqrt{ 9 } } { 2 \times 1 } }$
$x = \dfrac { 7 \pm \sqrt{ \color{#FF6800}{ 9 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$x = \dfrac { 7 \pm \color{#FF6800}{ 3 } } { 2 \times 1 }$
$x = \dfrac { 7 \pm 3 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { 7 \pm 3 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 \pm 3 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 + 3 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 7 - 3 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 7 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } } { 2 } \\ x = \dfrac { 7 - 3 } { 2 } \end{array}$
 Add $7$ and $3$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { 7 - 3 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 10 } { 2 } } \\ x = \dfrac { 7 - 3 } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 7 - 3 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 7 - 3 } { 2 } \end{array}$
 Reduce the fraction to the lowest term 
$\begin{array} {l} x = \color{#FF6800}{ 5 } \\ x = \dfrac { 7 - 3 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 7 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 2 } \end{array}$
 Subtract $3$ from $7$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 4 } } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 4 } { 2 } } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 2 } { 1 } } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { 2 } { 1 } } \end{array}$
 Reduce the fraction to the lowest term 
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ 2 } \end{array}$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 7 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 10$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 7 ^ { 2 } - 4 \times 1 \times 10$
$D = \color{#FF6800}{ 7 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 10$
 Calculate power 
$D = \color{#FF6800}{ 49 } - 4 \times 1 \times 10$
$D = 49 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 10$
 Multiplying any number by 1 does not change the value 
$D = 49 - 4 \times 10$
$D = 49 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 }$
 Multiply $- 4$ and $10$
$D = 49 \color{#FF6800}{ - } \color{#FF6800}{ 40 }$
$D = \color{#FF6800}{ 49 } \color{#FF6800}{ - } \color{#FF6800}{ 40 }$
 Subtract $40$ from $49$
$D = \color{#FF6800}{ 9 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 9 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = 7 , \alpha \beta = 10$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ - } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 Change the symbols of both sides of the equation 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 7 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { 10 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 7 } { 1 } } , \alpha \beta = \dfrac { 10 } { 1 }$
 Solve the sign of a fraction with a negative sign 
$\alpha + \beta = \color{#FF6800}{ \dfrac { 7 } { 1 } } , \alpha \beta = \dfrac { 10 } { 1 }$
$\alpha + \beta = \dfrac { 7 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 10 } { 1 }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = \color{#FF6800}{ 7 } , \alpha \beta = \dfrac { 10 } { 1 }$
$\alpha + \beta = 7 , \alpha \beta = \dfrac { 10 } { \color{#FF6800}{ 1 } }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = 7 , \alpha \beta = \color{#FF6800}{ 10 }$
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