Calculator search results

Formula
Solve the quadratic equation
Answer
circle-check-icon
expand-arrow-icon
Number of solution
Answer
circle-check-icon
expand-arrow-icon
Relationship between roots and coefficients
Answer
circle-check-icon
Graph
$y = x ^ { 2 } - 6 x - 112$
$y = 0$
$x$-intercept
$\left ( - 8 , 0 \right )$, $\left ( 14 , 0 \right )$
$y$-intercept
$\left ( 0 , - 112 \right )$
Minimum
$\left ( 3 , - 121 \right )$
Standard form
$y = \left ( x - 3 \right ) ^ { 2 } - 121$
$x ^{ 2 } -6x-112 = 0$
$\begin{array} {l} x = 14 \\ x = - 8 \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 14 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 14 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 14 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 14 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 14 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$
$\begin{array} {l} x = 14 \\ x = - 8 \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 112 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - 3 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 112 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - 3 \right ) ^ { 2 } = \color{#FF6800}{ 112 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x - 3 \right ) ^ { 2 } = 112 + \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } }$
$ $ Calculate power $ $
$\left ( x - 3 \right ) ^ { 2 } = 112 + \color{#FF6800}{ 9 }$
$\left ( x - 3 \right ) ^ { 2 } = \color{#FF6800}{ 112 } \color{#FF6800}{ + } \color{#FF6800}{ 9 }$
$ $ Add $ 112 $ and $ 9$
$\left ( x - 3 \right ) ^ { 2 } = \color{#FF6800}{ 121 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 121 }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \pm \sqrt{ \color{#FF6800}{ 121 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \pm \sqrt{ \color{#FF6800}{ 121 } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 11 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 11 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 11 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 11 } \\ x = 3 - 11 \end{array}$
$ $ Add $ 3 $ and $ 11$
$\begin{array} {l} x = \color{#FF6800}{ 14 } \\ x = 3 - 11 \end{array}$
$\begin{array} {l} x = 14 \\ x = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \end{array}$
$ $ Subtract $ 11 $ from $ 3$
$\begin{array} {l} x = 14 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$
$\begin{array} {l} x = 14 \\ x = - 8 \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 6 \right ) \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 112 \right ) } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 6 \pm \sqrt{ \left ( - 6 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 112 \right ) } } { 2 \times 1 }$
$x = \dfrac { 6 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 112 \right ) } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 1 \times \left ( - 112 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm \sqrt{ 6 ^ { 2 } - 4 \times 1 \times \left ( - 112 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm \sqrt{ 484 } } { 2 \times 1 } }$
$x = \dfrac { 6 \pm \sqrt{ \color{#FF6800}{ 484 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 6 \pm \color{#FF6800}{ 22 } } { 2 \times 1 }$
$x = \dfrac { 6 \pm 22 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 6 \pm 22 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 \pm 22 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 + 22 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 6 - 22 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 22 } } { 2 } \\ x = \dfrac { 6 - 22 } { 2 } \end{array}$
$ $ Add $ 6 $ and $ 22$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 28 } } { 2 } \\ x = \dfrac { 6 - 22 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 28 } { 2 } } \\ x = \dfrac { 6 - 22 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 14 } { 1 } } \\ x = \dfrac { 6 - 22 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 14 } { 1 } } \\ x = \dfrac { 6 - 22 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 14 } \\ x = \dfrac { 6 - 22 } { 2 } \end{array}$
$\begin{array} {l} x = 14 \\ x = \dfrac { \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 22 } } { 2 } \end{array}$
$ $ Subtract $ 22 $ from $ 6$
$\begin{array} {l} x = 14 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 16 } } { 2 } \end{array}$
$\begin{array} {l} x = 14 \\ x = \color{#FF6800}{ \dfrac { - 16 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 14 \\ x = \color{#FF6800}{ \dfrac { - 8 } { 1 } } \end{array}$
$\begin{array} {l} x = 14 \\ x = \dfrac { - 8 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 14 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 112 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 112 \right )$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 6 ^ { 2 } - 4 \times 1 \times \left ( - 112 \right )$
$D = \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 112 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 36 } - 4 \times 1 \times \left ( - 112 \right )$
$D = 36 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 112 \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 36 - 4 \times \left ( - 112 \right )$
$D = 36 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 112 } \right )$
$ $ Multiply $ - 4 $ and $ - 112$
$D = 36 + \color{#FF6800}{ 448 }$
$D = \color{#FF6800}{ 36 } \color{#FF6800}{ + } \color{#FF6800}{ 448 }$
$ $ Add $ 36 $ and $ 448$
$D = \color{#FF6800}{ 484 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 484 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = 6 , \alpha \beta = - 112$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 112 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 6 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 112 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 6 } { 1 } } , \alpha \beta = \dfrac { - 112 } { 1 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 6 } { 1 } } , \alpha \beta = \dfrac { - 112 } { 1 }$
$\alpha + \beta = \dfrac { 6 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 112 } { 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = \color{#FF6800}{ 6 } , \alpha \beta = \dfrac { - 112 } { 1 }$
$\alpha + \beta = 6 , \alpha \beta = \dfrac { - 112 } { \color{#FF6800}{ 1 } }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\alpha + \beta = 6 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 112 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$x^{2}$ $-6x-27=0$ 
$x^{2}-25=0$ 
$x^{2+10x+25}=0$ 
$2x^{2}-5x+3=0$
7th-9th grade
Other
search-thumbnail-$x^{2}-6x-11=-4$
10th-13th grade
Other
search-thumbnail-$x^{2}-x$ $-2=0$ 
$x^{2}+16x-36=0$ 
$8x^{2}-6x-5=0$ 
$x^{2}-7x+12=0$ 
$3x^{2}+5x+2=0$
7th-9th grade
Algebra
search-thumbnail-R-6x-120,findx?tL 
$x^{2}-6x-1=0$ $x^{2}+\dfrac {1} {x^{2}}$
7th-9th grade
Algebra
Have you found the solution you wanted?
Try again
Try more features at QANDA!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture
apple logogoogle play logo