Answer
$x=\frac{3-\sqrt {29}}{2},\frac{3+\sqrt {29}}{2}$
Work Step by Step
To complete the square, add a term that is the square of half the coefficient of $x$ to both sides of the equation, and factor the left side. Then use the square root property to simplify both sides, and then isolate $x$.
$x^2-3x-5=0$
$(\frac{-3}{2})^2$
$x^2-3x-5+5+(\frac{-3}{2})^2=5+(\frac{-3}{2})^2$
$x^2-3x+(\frac{-3}{2})^2=5+(\frac{-3}{2})^2$
$x^2-3x+\frac{9}{4}=5+\frac{9}{4}$
$x^2-3x+\frac{9}{4}=\frac{29}{4}$
$(x-\frac{3}{2})^2=\frac{29}{4}$
$x-\frac{3}{2}=±\frac{\sqrt {29}}{2}$
$x=\frac{3}{2}±\frac{\sqrt {29}}{2}$
$x=\frac{3-\sqrt {29}}{2},\frac{3+\sqrt {29}}{2}$
