# How to solve 5x^2+20x+13=0 by completing the square?

Mar 1, 2018

$x = \frac{- 10 \pm \sqrt{35}}{5}$

#### Explanation:

$5 {x}^{2} + 20 x + 13 = 0$
Multiplying both sides by 5, we get
$25 {x}^{2} + 100 x + 65 = 0$
$\Rightarrow 25 {x}^{2} + 2 \left(5 x\right) \left(10\right) + 100 - 35 = 0$
$\Rightarrow {\left(5 x\right)}^{2} + 2 \left(5 x\right) \left(10\right) + {\left(10\right)}^{2} - 35 = 0$
$\Rightarrow {\left(5 x + 10\right)}^{2} - 35 = 0$
$\Rightarrow {\left(5 x + 10\right)}^{2} = 35 = {\left(\sqrt{35}\right)}^{2}$
$\Rightarrow 5 x + 10 = \pm \sqrt{35}$
$5 x = - 10 \pm \sqrt{35}$
$x = \frac{- 10 \pm \sqrt{35}}{5}$

Mar 2, 2018

$5 {x}^{2} + 20 x + 13 = 0$

$\frac{1}{5} \left({x}^{2} + 4 x + \frac{13}{5}\right) = 0$

${x}^{2} + 4 x + \frac{13}{5} = 0$

Now, add and subtract, half of the x-term squared, i.e. ${\left(\frac{4}{2}\right)}^{2} = {2}^{2} = 4$

$\implies \left[{x}^{2} + 4 x + 4\right] + \left[\frac{13}{5} - 4\right] = 0$

$\implies \left[{x}^{2} + 4 x + {2}^{2}\right] = \left[4 - \frac{13}{5}\right]$

$\implies {\left(x + 2\right)}^{2} = \frac{20}{5} - \frac{13}{5}$

=>(x+2) =+-sqrt(7/5

=>x =-2+-sqrt(7/5