# How do you solve x^2+14x + 49= 19?

May 22, 2018

${x}^{2} + 14 x + 49 = 19$ has the two solutions $x = - 7 \pm \setminus \sqrt{19}$
(i.e. ${x}_{1} = - 7 - \setminus \sqrt{19}$ and ${x}_{2} = - 7 + \setminus \sqrt{19}$)

#### Explanation:

The first I notice is that $14 = 2 \cdot 7$, and $49 = {7}^{2}$

Please also remember that ${\left(x + a\right)}^{2} = {x}^{2} + 2 a + {a}^{2}$
Therefore ${x}^{2} + 14 x + 49 = {\left(x + 7\right)}^{2} = 19$

It follows, therefore, that
$x + 7 = \pm \setminus \sqrt{19}$

We, therefore, have two values of x here:
$x = - 7 \pm \setminus \sqrt{19}$