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

Mar 30, 2015

Group the two variable terms together, then factor out the number in front of ${x}^{2}$

$\left(2 {x}^{2} + 7 x\right) - 15 = 0$

$2 \left({x}^{2} + \frac{7}{2} x + \textcolor{w h i t e}{\text{leave space}}\right) - 15 = 0$

Take $\frac{1}{2}$ of the coefficient of $x$, square that and add and subtract inside the parentheses:

$\frac{1}{2} \text{ of } \frac{7}{2} = \frac{7}{4}$ Squaring gives us $\frac{49}{16}$, so we write:

$2 \left({x}^{2} + \frac{7}{2} x + \frac{49}{16} - \frac{49}{16}\right) - 15 = 0$ Keep the positive $\frac{49}{16}$ inside the parentheses (we need it for the perfect square)

Write:
$2 \left({x}^{2} + \frac{7}{2} x + \frac{49}{16}\right) - 2 \left(\frac{49}{16}\right) - 15 = 0$

Factor the square and simplify the rest:

$2 {\left(x + \frac{7}{4}\right)}^{2} - \frac{49}{8} - \frac{120}{8} = 0$

$2 {\left(x + \frac{7}{4}\right)}^{2} - \frac{169}{8} = 0$

$2 {\left(x + \frac{7}{4}\right)}^{2} = \frac{169}{8}$

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

$x + \frac{7}{4} = \pm \sqrt{\frac{169}{16}}$

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

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

$- \frac{7}{4} + \frac{13}{4} = \frac{6}{4} = \frac{3}{2}$ and $- \frac{7}{4} - \frac{13}{4} = - \frac{20}{4} = - 5$

The solutions are $\frac{3}{2}$, $5$