# How do you solve by completing the square: 2x^2 - 7x -15 = 0?

Jun 8, 2018

$x = - \frac{3}{2} \text{ or } x = 5$

#### Explanation:

$\text{to solve by "color(blue)"completing the square}$

$\text{add 15 to both sides}$

$2 {x}^{2} - 7 x = 15$

$\text{the coefficient of the "x^2" term must be 1}$

$\text{divide all terms by 2}$

${x}^{2} - \frac{7}{2} x = \frac{15}{2}$

$\text{add "(1/2"coefficient of x-term")^2" to both sides}$

${x}^{2} + 2 \left(- \frac{7}{4}\right) x \textcolor{red}{+ \frac{49}{16}} = \frac{15}{2} \textcolor{red}{+ \frac{49}{16}}$

${\left(x - \frac{7}{4}\right)}^{2} = \frac{169}{16}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$x - \frac{7}{4} = \pm \frac{13}{4}$

$x = \frac{7}{4} \pm \frac{13}{4}$

$x = \frac{7}{4} - \frac{13}{4} = - \frac{3}{2}$

$\text{or } x = \frac{7}{4} + \frac{13}{4} = 5$