How do you solve x^2 - 3x + 6 = 0 by completing the square?

Jul 5, 2017

$x = \frac{3}{2} \pm i \frac{\sqrt{15}}{2}$

Explanation:

${x}^{2} - 3 x + 6 = 0$

add and subtract half the coefficient of $x$ squared

$\textcolor{b l u e}{\left({x}^{2} - 3 x + {\left(\frac{3}{2}\right)}^{2}\right)} + 6 - {\left(\frac{3}{2}\right)}^{2} = 0$

the section in blue is a perfect square

$\textcolor{b l u e}{{\left(x - \frac{3}{2}\right)}^{2}} + \frac{15}{4} = 0$

solve by rearranging for $x$

${\left(x - \frac{3}{2}\right)}^{2} = - \frac{15}{4}$

$x - \frac{3}{2} = \sqrt{- \frac{15}{4}}$

$x - \frac{3}{2} = \pm i \frac{\sqrt{15}}{2}$

$x = \frac{3}{2} \pm i \frac{\sqrt{15}}{2}$