Answer
$x\approx 6.27$ and $x\approx -1.27$.
Work Step by Step
The given equation is
$\Rightarrow x^2-5x=8$
Find the value of $(\frac{b}{2})^2$.
Substitute $-5$ for $b$.
$=(\frac{-5}{2})^2$
Simplify.
$=\frac{25}{4}$
Add $\frac{25}{4}$ to each side of the equation.
$\Rightarrow x^2-5x+\frac{25}{4}=8+\frac{25}{4}$
Simplify.
$\Rightarrow x^2-5x+\frac{25}{4}=\frac{32+25}{4}$
$\Rightarrow x^2-5x+\frac{25}{4}=\frac{57}{4}$
Write the left side as the square of a binomial.
$\Rightarrow (x-\frac{5}{2})^2=\frac{57}{4}$
Take the square root of each side.
$\Rightarrow x-\frac{5}{2}=\pm \sqrt{\frac{57}{4}}$
Add $\frac{5}{2}$ from each side.
$\Rightarrow x-\frac{5}{2}+\frac{5}{2}=\pm \sqrt{\frac{57}{4}}+\frac{5}{2}$
Simplify.
$\Rightarrow x=\pm\frac{ \sqrt{57}}{2}+\frac{5}{2}$
The solutions are $x=\frac{ \sqrt{57}}{2}+\frac{5}{2}\approx 6.27$ and $x=-\frac{ \sqrt{57}}{2}+\frac{5}{2}\approx -1.27$.
