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

May 11, 2018

Solution: $x = 1.75 \pm \frac{1}{4} \sqrt{47} i$

#### Explanation:

$2 {x}^{2} - 7 x + 12 = 0 \mathmr{and} 2 {x}^{2} - 7 x = - 12$ or

$2 \left({x}^{2} - \frac{7}{2} x\right) = - 12$ or

$2 \left({x}^{2} - \frac{7}{2} x + \frac{49}{16}\right) = \frac{49}{8} - 12$ or

$2 {\left(x - \frac{7}{4}\right)}^{2} = - \frac{47}{8}$ or

${\left(x - \frac{7}{4}\right)}^{2} = - \frac{47}{16}$ or

$\left(x - \frac{7}{4}\right) = \pm \sqrt{- \frac{47}{16}}$ or

$\left(x - \frac{7}{4}\right) = \pm \frac{\sqrt{47 {i}^{2}}}{4} \left[{i}^{2} = - 1\right]$

$x = \frac{7}{4} \pm \frac{1}{4} \sqrt{47} i$ or

$x = 1.75 \pm \frac{1}{4} \sqrt{47} i$

Solution: $x = 1.75 \pm \frac{1}{4} \sqrt{47} i$ [Ans]