# How do you solve 2x^2-7x+12=0 by completing the square?

Nov 16, 2016

$x = \frac{7}{4} \pm \frac{i \sqrt{- 47}}{4}$

#### Explanation:

Given -

$2 {x}^{2} - 7 x + 12 = 0$

$2 {x}^{2} - 7 x = - 12$
Divide both sides by $2$
$\frac{\cancel{2} {x}^{2}}{\cancel{2}} - \frac{7}{2} x = \frac{- 12}{2}$
${x}^{2} - \frac{7}{2} x = - 6$
Divide$\frac{7}{2}$ by $2$, square it and add it to both sides

${x}^{2} - \frac{7}{2} \cdot \frac{1}{2} \cdot x + \frac{49}{16} = - 6 + \frac{49}{16}$
${x}^{2} - \frac{7}{4} + \frac{49}{16} = - 6 + \frac{49}{16}$
${x}^{2} - \frac{7}{4} + \frac{49}{16} = \frac{- 96 + 49}{16}$
${\left(x - \frac{7}{4}\right)}^{2} = \frac{- 47}{16}$
$x - \frac{7}{4} = \pm \sqrt{\frac{- 47}{16}}$
$x = \frac{7}{4} \pm \frac{i \sqrt{- 47}}{4}$