# How do you solve by completing the square: x^2 + 3x - 18 = 0?

Mar 30, 2015

In this way:

${x}^{2} + 3 x - 18 = 0 \Rightarrow {x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} - 18 = 0 \Rightarrow$

$\left({x}^{2} + 3 x + \frac{9}{4}\right) - \frac{9}{4} - 18 = 0 \Rightarrow {\left(x + \frac{3}{2}\right)}^{2} = 18 + \frac{9}{4} \Rightarrow$

${\left(x + \frac{3}{2}\right)}^{2} = \frac{72 + 9}{4} \Rightarrow {\left(x + \frac{3}{2}\right)}^{2} = \frac{81}{4} \Rightarrow$

$x + \frac{3}{2} = \pm \frac{9}{2} \Rightarrow x = - \frac{3}{2} \pm \frac{9}{2} \Rightarrow$

${x}_{1} = - \frac{3}{2} - \frac{9}{2} = - \frac{12}{2} = - 6$

${x}_{2} = - \frac{3}{2} + \frac{9}{2} = \frac{6}{2} = 3$.