# How do you solve x^2+10x=3 by completing the square?

Mar 27, 2016

See explanation...

#### Explanation:

Add ${\left(\frac{10}{2}\right)}^{2} = {5}^{2} = 25$ to both sides to get:

${x}^{2} + 10 x + 25 = 28$

The left hand side is equal to ${\left(x + 5\right)}^{2}$ so we have:

${\left(x + 5\right)}^{2} = 28 = {2}^{2} \cdot 7$

Hence:

$x + 5 = \pm \sqrt{{2}^{2} \cdot 7} = \pm 2 \sqrt{7}$

So:

$x = - 5 \pm 2 \sqrt{7}$

$\textcolor{w h i t e}{}$
$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$
From which we can derive the quadratic formula for zeros of $a {x}^{2} + b x + c$, namely:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$