# How do you solve -2x^2+10x=-14 by completing the square?

Jan 22, 2018

−2x^2+10x=−14

$2 {x}^{2} - 10 x - 14 = 0$

dividing the entire equation by 2,
${x}^{2} - 5 x - 7 = 0$

now adding and subtract "half of the $x$ term squared",
which in this case is, ${\left(\frac{5}{2}\right)}^{2}$

$\left({x}^{2} - 5 x + {\left(\frac{5}{2}\right)}^{2}\right) - 7 - {\left(\frac{5}{2}\right)}^{2} = 0$

${\left(x - \frac{5}{2}\right)}^{2} - 7 - \left(\frac{25}{4}\right) = 0$

${\left(x - \frac{5}{2}\right)}^{2} - \frac{28}{4} - \frac{25}{4} = 0$

${\left(x - \frac{5}{2}\right)}^{2} - \frac{53}{4} = 0$

${\left(x - \frac{5}{2}\right)}^{2} = \frac{53}{4}$

$\left(x - \frac{5}{2}\right) = \pm \sqrt{\frac{53}{4}}$

$\left(x - \frac{5}{2}\right) = \pm \frac{\sqrt{53}}{2}$

$x = \pm \frac{\sqrt{53}}{2} + \frac{5}{2}$

$x = \frac{5 \pm \sqrt{53}}{2}$