qanda-logo
apple logogoogle play logo

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
expand-arrow-icon
Graph
$y = 6 x ^ { 2 } - 11 x - 10$
$y = 0$
$x$Intercept
$\left ( - \dfrac { 2 } { 3 } , 0 \right )$, $\left ( \dfrac { 5 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 10 \right )$
Minimum
$\left ( \dfrac { 11 } { 12 } , - \dfrac { 361 } { 24 } \right )$
Standard form
$y = 6 \left ( x - \dfrac { 11 } { 12 } \right ) ^ { 2 } - \dfrac { 361 } { 24 }$
$6x ^{ 2 } -11x-10 = 0$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = - \dfrac { 2 } { 3 } \end{array}$
Find solution by method of factorization
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = 0$
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$
$ $ If the product of the factor is 0, at least one factor should be 0 $ $
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$
$ $ Solve the equation to find $ x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = - \dfrac { 2 } { 3 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ Divide both sides by the coefficient of the leading highest term $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 6 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 6 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } = \color{#FF6800}{ 0 }$
$ $ Convert the quadratic expression on the left side to a perfect square format $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 11 } { 12 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
$ $ Move the constant to the right side and change the sign $ $
$\left ( x - \dfrac { 11 } { 12 } \right ) ^ { 2 } = \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 11 } { 12 } \right ) ^ { 2 } = \dfrac { 5 } { 3 } + \left ( \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ When raising a fraction to the power, raise the numerator and denominator each to the power $ $
$\left ( x - \dfrac { 11 } { 12 } \right ) ^ { 2 } = \dfrac { 5 } { 3 } + \dfrac { \color{#FF6800}{ 11 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 12 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 ^ { 2 } } { 12 ^ { 2 } } }$
$ $ Organize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 361 } { 144 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 12 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 361 } { 144 } }$
$ $ Solve quadratic equations using the square root $ $
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 12 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 361 } { 144 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 11 } { 12 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 361 } { 144 } } }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 19 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 12 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 19 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 12 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 19 } { 12 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 19 } { 12 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 19 } { 12 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 19 } { 12 } } \end{array}$
$ $ Organize the expression $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = - \dfrac { 2 } { 3 } \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 11 \right ) \pm \sqrt{ \left ( - 11 \right ) ^ { 2 } - 4 \times 6 \times \left ( - 10 \right ) } } { 2 \times 6 }$
$ $ Simplify Minus $ $
$x = \dfrac { 11 \pm \sqrt{ \left ( - 11 \right ) ^ { 2 } - 4 \times 6 \times \left ( - 10 \right ) } } { 2 \times 6 }$
$x = \dfrac { 11 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 6 \times \left ( - 10 \right ) } } { 2 \times 6 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 11 \pm \sqrt{ 11 ^ { 2 } - 4 \times 6 \times \left ( - 10 \right ) } } { 2 \times 6 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 \pm \sqrt{ 11 ^ { 2 } - 4 \times 6 \times \left ( - 10 \right ) } } { 2 \times 6 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 \pm \sqrt{ 361 } } { 2 \times 6 } }$
$x = \dfrac { 11 \pm \sqrt{ \color{#FF6800}{ 361 } } } { 2 \times 6 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 11 \pm \color{#FF6800}{ 19 } } { 2 \times 6 }$
$x = \dfrac { 11 \pm 19 } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } }$
$ $ Multiply $ 2 $ and $ 6$
$x = \dfrac { 11 \pm 19 } { \color{#FF6800}{ 12 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 \pm 19 } { 12 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 + 19 } { 12 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 11 - 19 } { 12 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 11 } \color{#FF6800}{ + } \color{#FF6800}{ 19 } } { 12 } \\ x = \dfrac { 11 - 19 } { 12 } \end{array}$
$ $ Add $ 11 $ and $ 19$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 30 } } { 12 } \\ x = \dfrac { 11 - 19 } { 12 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 30 } { 12 } } \\ x = \dfrac { 11 - 19 } { 12 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ x = \dfrac { 11 - 19 } { 12 } \end{array}$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = \dfrac { \color{#FF6800}{ 11 } \color{#FF6800}{ - } \color{#FF6800}{ 19 } } { 12 } \end{array}$
$ $ Subtract $ 19 $ from $ 11$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } } { 12 } \end{array}$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = \color{#FF6800}{ \dfrac { - 8 } { 12 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = \color{#FF6800}{ \dfrac { - 2 } { 3 } } \end{array}$
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 3 } \end{array}$
$ $ Move the minus sign to the front of the fraction $ $
$\begin{array} {l} x = \dfrac { 5 } { 2 } \\ x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \end{array}$
$ $ 2 real roots $ $
Find the number of solutions
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 11 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 6 \times \left ( - 10 \right )$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 11 ^ { 2 } - 4 \times 6 \times \left ( - 10 \right )$
$D = \color{#FF6800}{ 11 } ^ { \color{#FF6800}{ 2 } } - 4 \times 6 \times \left ( - 10 \right )$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 121 } - 4 \times 6 \times \left ( - 10 \right )$
$D = 121 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \right )$
$ $ Multiply the numbers $ $
$D = 121 + \color{#FF6800}{ 240 }$
$D = \color{#FF6800}{ 121 } \color{#FF6800}{ + } \color{#FF6800}{ 240 }$
$ $ Add $ 121 $ and $ 240$
$D = \color{#FF6800}{ 361 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 361 }$
$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $
$ $ 2 real roots $ $
$\alpha + \beta = \dfrac { 11 } { 6 } , \alpha \beta = - \dfrac { 5 } { 3 }$
Find the sum and product of the two roots of the quadratic equation
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 11 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } = \color{#FF6800}{ 0 }$
$ $ 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}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 11 } { 6 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 10 } { 6 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 11 } { 6 } } , \alpha \beta = \dfrac { - 10 } { 6 }$
$ $ Solve the sign of a fraction with a negative sign $ $
$\alpha + \beta = \color{#FF6800}{ \dfrac { 11 } { 6 } } , \alpha \beta = \dfrac { - 10 } { 6 }$
$\alpha + \beta = \dfrac { 11 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { - 10 } { 6 } }$
$ $ Reduce the fraction $ $
$\alpha + \beta = \dfrac { 11 } { 6 } , \alpha \beta = \color{#FF6800}{ \dfrac { - 5 } { 3 } }$
$\alpha + \beta = \dfrac { 11 } { 6 } , \alpha \beta = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 5 } } { 3 }$
$ $ Move the minus sign to the front of the fraction $ $
$\alpha + \beta = \dfrac { 11 } { 6 } , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-PRACTICE QUESTIONS 
CLASS X : CHAPTER - 4 
$vADRAx1C$ EQUATIONS 
FACTORISATION METHOD 
Solve the following quadratic equations: 
1. $x^{2}+11x+30=0$ $2430x^{2}+7x-15=0$ 
+ 
2. $x^{2}+18x+32=0$ $25.24x^{2}-41x+12=0$ 
- 
3. $x^{2}+7x-18=0$ $26.2x^{2}-7x-15=0$ 
- - 
4. x² + 5x 6 - 0 $27.6x^{2}+11x-10=0$ 
– - 
5. y- 4y + 3 =0 $28.10x^{2}-9x-7=0$ 
- 
6. $x^{2}-21x+108=0$ $29.5x^{2}-16x-21=0$ 
- – 
7. $x^{2}-11x-80=0$ $30.2x^{2}-x-21=0$ 
8. x-x- 156 -0 $31.15x^{2}-x-28=0$ 
- 
9. $x^{2}-32x-105=0$ $32.8x^{2}-27b+9b^{2}=0$ 
- 
$1040+3x-x^{2}=0$ $33.5x^{2}+33xy-14y^{2}=0$ 
– –
10th-13th grade
Algebra
search-thumbnail-Factorise: $x^{2}-11x-102$
7th-9th grade
Other
search-thumbnail-$3$ What is the standard form of the inequality 
$\left(2x+3\right)\left(x+4\right)>8x^{2}-27$ 
$A$ $6x^{2}+11x+14>0$ 
$B$ $6x^{2}+11x-10>0$ 
C. $6x^{2}-11x-14<0$ 
D. $6x^{2}-11x-10<0$
10th-13th grade
Other
search-thumbnail-$6$ $x^{2}$ $+5x$ $-$ $176$ 
$10$ $x^{2}-11x$ $-102$
10th-13th grade
Other
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