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Formula
Solve the quadratic equation
Number of solution
Relationship between roots and coefficients
$x ^{ 2 } -5x-6=0x$
$\begin{array} {l} x = 6 \\ x = - 1 \end{array}$
Find solution by method of factorization
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 } \color{#FF6800}{ x }$
 Move the expression to the left side and change the symbol 
$x ^ { 2 } - 5 x - 6 \color{#FF6800}{ - } \color{#FF6800}{ 0 } \color{#FF6800}{ x } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 0 } \color{#FF6800}{ x } = 0$
 Expand the expression 
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 0 }$
 If the product of the factor is 0, at least one factor should be 0 
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \end{array}$
 Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = 6 \\ x = - 1 \end{array}$
Solve quadratic equations using the square root
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 } \color{#FF6800}{ x }$
 Organize the expression 
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
 Convert the quadratic expression on the left side to a perfect square format 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
 Move the constant to the right side and change the sign 
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = \color{#FF6800}{ 6 } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = 6 + \left ( \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
 When raising a fraction to the power, raise the numerator and denominator each to the power 
$\left ( x - \dfrac { 5 } { 2 } \right ) ^ { 2 } = 6 + \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 ^ { 2 } } { 2 ^ { 2 } } }$
 Organize the expression 
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 49 } { 4 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 49 } { 4 } }$
 Solve quadratic equations using the square root 
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 49 } { 4 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 49 } { 4 } } }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 7 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 2 } } \end{array}$
 Organize the expression 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = 6 \\ x = - 1 \end{array}$
Calculate using the quadratic formula
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 } \color{#FF6800}{ x }$
 Organize the expression 
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 }$
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 5 \right ) \pm \sqrt{ \left ( - 5 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 6 \right ) } } { 2 \times 1 }$
 Simplify Minus 
$x = \dfrac { 5 \pm \sqrt{ \left ( - 5 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 6 \right ) } } { 2 \times 1 }$
$x = \dfrac { 5 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 6 \right ) } } { 2 \times 1 }$
 Remove negative signs because negative numbers raised to even powers are positive 
$x = \dfrac { 5 \pm \sqrt{ 5 ^ { 2 } - 4 \times 1 \times \left ( - 6 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 \pm \sqrt{ 5 ^ { 2 } - 4 \times 1 \times \left ( - 6 \right ) } } { 2 \times 1 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 \pm \sqrt{ 49 } } { 2 \times 1 } }$
$x = \dfrac { 5 \pm \sqrt{ \color{#FF6800}{ 49 } } } { 2 \times 1 }$
 Organize the part that can be taken out of the radical sign inside the square root symbol 
$x = \dfrac { 5 \pm \color{#FF6800}{ 7 } } { 2 \times 1 }$
$x = \dfrac { 5 \pm 7 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
 Multiplying any number by 1 does not change the value 
$x = \dfrac { 5 \pm 7 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 \pm 7 } { 2 } }$
 Separate the answer 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 + 7 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 - 7 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ + } \color{#FF6800}{ 7 } } { 2 } \\ x = \dfrac { 5 - 7 } { 2 } \end{array}$
 Add $5$ and $7$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 12 } } { 2 } \\ x = \dfrac { 5 - 7 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 12 } { 2 } } \\ x = \dfrac { 5 - 7 } { 2 } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 } { 1 } } \\ x = \dfrac { 5 - 7 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 6 } { 1 } } \\ x = \dfrac { 5 - 7 } { 2 } \end{array}$
 Reduce the fraction to the lowest term 
$\begin{array} {l} x = \color{#FF6800}{ 6 } \\ x = \dfrac { 5 - 7 } { 2 } \end{array}$
$\begin{array} {l} x = 6 \\ x = \dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 7 } } { 2 } \end{array}$
 Subtract $7$ from $5$
$\begin{array} {l} x = 6 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 2 } \end{array}$
$\begin{array} {l} x = 6 \\ x = \color{#FF6800}{ \dfrac { - 2 } { 2 } } \end{array}$
 Do the reduction of the fraction format 
$\begin{array} {l} x = 6 \\ x = \color{#FF6800}{ \dfrac { - 1 } { 1 } } \end{array}$
$\begin{array} {l} x = 6 \\ x = \dfrac { - 1 } { \color{#FF6800}{ 1 } } \end{array}$
 If the denominator is 1, the denominator can be removed 
$\begin{array} {l} x = 6 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
 2 real roots 
Find the number of solutions
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 } \color{#FF6800}{ x }$
 Organize the expression 
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right )$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 6 \right )$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 5 ^ { 2 } - 4 \times 1 \times \left ( - 6 \right )$
$D = \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 6 \right )$
 Calculate power 
$D = \color{#FF6800}{ 25 } - 4 \times 1 \times \left ( - 6 \right )$
$D = 25 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 6 \right )$
 Multiplying any number by 1 does not change the value 
$D = 25 - 4 \times \left ( - 6 \right )$
$D = 25 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right )$
 Multiply $- 4$ and $- 6$
$D = 25 + \color{#FF6800}{ 24 }$
$D = \color{#FF6800}{ 25 } \color{#FF6800}{ + } \color{#FF6800}{ 24 }$
 Add $25$ and $24$
$D = \color{#FF6800}{ 49 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 49 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
$\alpha + \beta = 5 , \alpha \beta = - 6$
Find the sum and product of the two roots of the quadratic equation
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 } \color{#FF6800}{ x }$
 Organize the expression 
$x ^ { 2 } - 5 x - 6 = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ 0 }$
 In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 5 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 6 } { 1 } }$
$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 5 } { 1 } } , \alpha \beta = \dfrac { - 6 } { 1 }$
 Solve the sign of a fraction with a negative sign 
$\alpha + \beta = \color{#FF6800}{ \dfrac { 5 } { 1 } } , \alpha \beta = \dfrac { - 6 } { 1 }$
$\alpha + \beta = \dfrac { 5 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 6 } { 1 }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = \color{#FF6800}{ 5 } , \alpha \beta = \dfrac { - 6 } { 1 }$
$\alpha + \beta = 5 , \alpha \beta = \dfrac { - 6 } { \color{#FF6800}{ 1 } }$
 If the denominator is 1, the denominator can be removed 
$\alpha + \beta = 5 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 6 }$
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