# How do you solve by completing the square: x^2+3x-4=0?

Apr 3, 2015

x=1 , -4
Take half of the coefficient of x, that would be $\frac{3}{2}$ in this case. Square this number and add it and subtract it as shown below.

${x}^{2}$ +3x + $\frac{9}{4}$ -$\frac{9}{4}$-4=0

${\left(x + \frac{3}{2}\right)}^{2}$= $\frac{25}{4}$. Take square root of both sides to have x+$\frac{3}{2}$ = $\frac{5}{2}$ or $- \frac{5}{2}$

x+$\frac{3}{2}$ = $\frac{5}{2}$ would give x= $\frac{5}{2}$ -$\frac{3}{2}$ = 1 and

X+$\frac{3}{2}$ = $- \frac{5}{2}$ would give x= $- \frac{5}{2}$ $- \frac{3}{2}$ = -4