Answer
$x=\displaystyle \frac{3}{2}+\frac{\sqrt{29}}{2}$ or $x=\displaystyle \frac{3}{2}-\frac{\sqrt{29}}{2}$
Work Step by Step
$ x^{2}=3x+5\qquad$...add $-3x-5$ to both sides.
$x^{2}-3x-5=3x+5-3x-5$
$ x^{2}-3x-5=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-3,\ c=-5$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-3,\ a$ for $1$ and $c$ for $-5$.
$ x=\displaystyle \frac{-(-3)\pm\sqrt{(-3)^{2}-4\cdot(-5)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$x=\displaystyle \frac{3\pm\sqrt{9+20}}{2}$
$ x=\displaystyle \frac{3\pm\sqrt{29}}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$x=\displaystyle \frac{3}{2}+\frac{\sqrt{29}}{2}$ or $x=\displaystyle \frac{3}{2}-\frac{\sqrt{29}}{2}$
