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

Apr 5, 2017

${\left(x + \frac{7}{2}\right)}^{2} + \frac{11}{4} = 0$

#### Explanation:

The process of completing the square yields the standard form or
Vertex form $y = a {\left(x - h\right)}^{2} + k$, where vertex: $\left(h , k\right)$

Use completing of the square by grouping the $x$-terms:
$\left({x}^{2} + 7 x\right) + 15 = 0$

Get the squared value by halving the $x$-term: $\frac{1}{2} \cdot 7 = \frac{7}{2}$
${\left(x + \frac{7}{2}\right)}^{2} + 15 - {\left(\frac{7}{2}\right)}^{2} = 0$

You need to subtract ${\left(\frac{7}{2}\right)}^{2} = \frac{49}{4}$ because it is added when the square is completed:

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

From $\text{ } {\left(x + \frac{7}{2}\right)}^{2} + 15 - \frac{49}{4} = 0$

$\text{ } {\left(x + \frac{7}{2}\right)}^{2} + \frac{60}{4} - \frac{49}{4} = 0$

${\left(x + \frac{7}{2}\right)}^{2} + \frac{11}{4} = 0$