We'll use the to solve the given .
a x 2 + b x + c = 0 ⇔ x = - b ± b 2 − 4 a c 2 a \begin{gathered}
{\color{#0000FF}{a}}x^2+{\color{#FF0000}{b}}x+{\color{#009600}{c}}=0 \quad \Leftrightarrow \quad x=\dfrac{\text{-} {\color{#FF0000}{b}}\pm \sqrt{{\color{#FF0000}{b}}^2-4{\color{#0000FF}{a}}{\color{#009600}{c}}}}{2{\color{#0000FF}{a}}}
\end{gathered} a x 2 + b x + c = 0 ⇔ x = 2 a - b ± b 2 − 4 a c
We first need to identify the values of
a , {\color{#0000FF}{a}}, a , b , {\color{#FF0000}{b}}, b , and
c . {\color{#009600}{c}}. c . 2 x 2 + ( - 16 ) x + 50 = 0 \begin{gathered}
{\color{#0000FF}{2}}x^2+({\color{#FF0000}{\text{-} 16}})x+{\color{#009600}{50}}=0
\end{gathered} 2 x 2 + ( - 1 6 ) x + 5 0 = 0
We see that
a = 2 , {\color{#0000FF}{a}}={\color{#0000FF}{2}}, a = 2 , b = - 16 , {\color{#FF0000}{b}}={\color{#FF0000}{\text{-} 16}}, b = - 1 6 , and
c = 50 . {\color{#009600}{c}}={\color{#009600}{50}}. c = 5 0 . Let's substitute these values into the Quadratic Formula.
x = - b ± b 2 − 4 a c 2 a x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a} x = 2 a - b ± b 2 − 4 a c x = - ( - 16 ) ± ( - 16 ) 2 − 4 ( 2 ) ( 50 ) 2 ( 2 ) x=\dfrac{\text{-} ({\color{#FF0000}{\text{-} 16}})\pm\sqrt{({\color{#FF0000}{\text{-} 16}})^2-4({\color{#0000FF}{2}})({\color{#009600}{50}})}}{2({\color{#0000FF}{2}})} x = 2 ( 2 ) - ( - 1 6 ) ± ( - 1 6 ) 2 − 4 ( 2 ) ( 5 0 )
Solve for
x x x and Simplify
x = 16 ± ( - 16 ) 2 − 4 ( 2 ) ( 50 ) 2 ( 2 ) x=\dfrac{16\pm\sqrt{(\text{-} 16)^2-4(2)(50)}}{2(2)} x = 2 ( 2 ) 1 6 ± ( - 1 6 ) 2 − 4 ( 2 ) ( 5 0 ) x = 16 ± 256 − 4 ( 2 ) ( 50 ) 2 ( 2 ) x=\dfrac{16\pm\sqrt{256-4(2)(50)}}{2(2)} x = 2 ( 2 ) 1 6 ± 2 5 6 − 4 ( 2 ) ( 5 0 ) x = 16 ± 256 − 4 ( 100 ) 4 x=\dfrac{16\pm\sqrt{256-4(100)}}{4} x = 4 1 6 ± 2 5 6 − 4 ( 1 0 0 ) x = 16 ± 256 − 400 4 x=\dfrac{16\pm\sqrt{256-400}}{4} x = 4 1 6 ± 2 5 6 − 4 0 0 x = 16 ± - 144 4 x=\dfrac{16\pm\sqrt{\text{-} 144}}{4} x = 4 1 6 ± - 1 4 4 x = 16 ± - 1 ⋅ 144 4 x=\dfrac{16\pm\sqrt{\text{-} 1 \cdot 144}}{4} x = 4 1 6 ± - 1 ⋅ 1 4 4 x = 16 ± - 1 ⋅ 144 4 x=\dfrac{16 \pm \sqrt{\text{-} 1} \cdot \sqrt{144}}{4} x = 4 1 6 ± - 1 ⋅ 1 4 4 x = 16 ± i ⋅ 144 4 x=\dfrac{16 \pm i \cdot \sqrt{144}}{4} x = 4 1 6 ± i ⋅ 1 4 4 x = 16 ± i ⋅ 12 4 x=\dfrac{16 \pm i \cdot 12}{4} x = 4 1 6 ± i ⋅ 1 2 x = 4 ( 4 ± i ⋅ 3 ) 4 x=\dfrac{4(4 \pm i \cdot 3)}{4} x = 4 4 ( 4 ± i ⋅ 3 ) x = 4 ± i ⋅ 3 x=4 \pm i \cdot3 x = 4 ± i ⋅ 3
Using the Quadratic Formula, we found that the solutions of the given equation are
x = 4 ± 3 i . x=4 \pm 3i. x = 4 ± 3 i . Thus, the solutions are
x 1 = 4 + 3 i x_1=4 + 3i x 1 = 4 + 3 i and
x 2 = 4 − 3 i . x_2=4- 3i. x 2 = 4 − 3 i . 