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$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 16 } = 0$

$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right ) = 0$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right ) = \color{#FF6800}{ 0 }$

$ $ If the product of the factor is 0, at least one factor should be 0 $ $

$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } = \color{#FF6800}{ 0 } \end{array}$

$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } = \color{#FF6800}{ 0 } \end{array}$

$ $ Solve the equation to find $ x$

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 16 } = \color{#FF6800}{ 0 }$

$ $ Convert the quadratic expression on the left side to a perfect square format $ $

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 16 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$

$\left ( x + 3 \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 16 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } = 0$

$ $ Move the constant to the right side and change the sign $ $

$\left ( x + 3 \right ) ^ { 2 } = \color{#FF6800}{ 16 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } }$

$\left ( x + 3 \right ) ^ { 2 } = 16 + \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } }$

$ $ Calculate power $ $

$\left ( x + 3 \right ) ^ { 2 } = 16 + \color{#FF6800}{ 9 }$

$\left ( x + 3 \right ) ^ { 2 } = \color{#FF6800}{ 16 } \color{#FF6800}{ + } \color{#FF6800}{ 9 }$

$ $ Add $ 16 $ and $ 9$

$\left ( x + 3 \right ) ^ { 2 } = \color{#FF6800}{ 25 }$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 25 }$

$ $ Solve quadratic equations using the square root $ $

$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \pm \sqrt{ \color{#FF6800}{ 25 } }$

$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } = \pm \sqrt{ \color{#FF6800}{ 25 } }$

$ $ Solve a solution to $ x$

$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 3 }$

$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 3 }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \\ x = - 3 - 5 \end{array}$

$ $ Add $ - 3 $ and $ 5$

$\begin{array} {l} x = \color{#FF6800}{ 2 } \\ x = - 3 - 5 \end{array}$

$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$

$ $ Find the sum of the negative numbers $ $

$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 16 } = \color{#FF6800}{ 0 }$

$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \pm \sqrt{ \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 16 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \pm \sqrt{ \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 16 } \right ) } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$

$ $ Organize the expression $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \pm \sqrt{ \color{#FF6800}{ 100 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$

$x = \dfrac { - 6 \pm \sqrt{ \color{#FF6800}{ 100 } } } { 2 \times 1 }$

$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $

$x = \dfrac { - 6 \pm \color{#FF6800}{ 10 } } { 2 \times 1 }$

$x = \dfrac { - 6 \pm 10 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$

$ $ Multiplying any number by 1 does not change the value $ $

$x = \dfrac { - 6 \pm 10 } { \color{#FF6800}{ 2 } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \pm \color{#FF6800}{ 10 } } { \color{#FF6800}{ 2 } } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { \color{#FF6800}{ 2 } } } \end{array}$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { - 6 - 10 } { 2 } \end{array}$

$ $ Add $ - 6 $ and $ 10$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 4 } } { 2 } \\ x = \dfrac { - 6 - 10 } { 2 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 2 } } } \\ x = \dfrac { - 6 - 10 } { 2 } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { - 6 - 10 } { 2 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { - 6 - 10 } { 2 } \end{array}$

$ $ Reduce the fraction to the lowest term $ $

$\begin{array} {l} x = \color{#FF6800}{ 2 } \\ x = \dfrac { - 6 - 10 } { 2 } \end{array}$

$\begin{array} {l} x = 2 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 2 } \end{array}$

$ $ Find the sum of the negative numbers $ $

$\begin{array} {l} x = 2 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 16 } } { 2 } \end{array}$

$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 16 } } { \color{#FF6800}{ 2 } } } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } } { \color{#FF6800}{ 1 } } } \end{array}$

$\begin{array} {l} x = 2 \\ x = \dfrac { - 8 } { \color{#FF6800}{ 1 } } \end{array}$

$ $ If the denominator is 1, the denominator can be removed $ $

$\begin{array} {l} x = 2 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 8 } \end{array}$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 16 } = \color{#FF6800}{ 0 }$

$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$

$\color{#FF6800}{ D } = \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 16 } \right )$

$D = \color{#FF6800}{ 6 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 16 \right )$

$ $ Calculate power $ $

$D = \color{#FF6800}{ 36 } - 4 \times 1 \times \left ( - 16 \right )$

$D = 36 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 16 \right )$

$ $ Multiplying any number by 1 does not change the value $ $

$D = 36 - 4 \times \left ( - 16 \right )$

$D = 36 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 16 } \right )$

$ $ Multiply $ - 4 $ and $ - 16$

$D = 36 + \color{#FF6800}{ 64 }$

$D = \color{#FF6800}{ 36 } \color{#FF6800}{ + } \color{#FF6800}{ 64 }$

$ $ Add $ 36 $ and $ 64$

$D = \color{#FF6800}{ 100 }$

$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$

$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $

$ $ 2 real roots $ $

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 16 } = \color{#FF6800}{ 0 }$

$ $ 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}$

$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 1 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 16 } } { \color{#FF6800}{ 1 } } }$

$\alpha + \beta = - \dfrac { 6 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 16 } { 1 }$

$ $ If the denominator is 1, the denominator can be removed $ $

$\alpha + \beta = - \color{#FF6800}{ 6 } , \alpha \beta = \dfrac { - 16 } { 1 }$

$\alpha + \beta = - 6 , \alpha \beta = \dfrac { - 16 } { \color{#FF6800}{ 1 } }$

$ $ If the denominator is 1, the denominator can be removed $ $

$\alpha + \beta = - 6 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 16 }$