Answer
$x=2$ and $x=-10$
Work Step by Step
$\dfrac{1}{8}x^{2}+x=\dfrac{5}{2}$
Multiply the whole equation by $8$ to avoid working with fractions:
$8\Big(\dfrac{1}{8}x^{2}+x=\dfrac{5}{2}\Big)$
$x^{2}+8x=20$
Take the $20$ to the left side of the equation:
$x^{2}+8x-20=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=8$ and $c=-20$
Substitute:
$x=\dfrac{-8\pm\sqrt{8^{2}-4(1)(-20)}}{2(1)}=\dfrac{-8\pm\sqrt{64+80}}{2}=...$
$...=\dfrac{-8\pm\sqrt{144}}{2}=\dfrac{-8\pm12}{2}$
Our two solutions are:
$x=\dfrac{-8+12}{2}=2$
$x=\dfrac{-8-12}{2}=-10$
