Question

In the following exercises, solve by completing the square.

$2n^2+4n=26$

Short answer

The solution is$n=-1-\sqrt{14},n=-1+\sqrt{14}$.

Step-by-step solution

Step 1. Given information 

Equation:$2n^2+4n=26$

Step 2. Calculation   

We have,

$2n^2+4n=26$

$2\left(n^2+2n\right)=2\left(13\right)$

$\frac{2\left(n^2+2n\right)}{2}=\frac{2\left(13\right)}{2}$

$n^2+2n=13$

role="math" localid="1645175362355" $n^2+2n+{\left(\frac{1}{2}·\left(2\right)\right)}^2=13+{\left(\frac{1}{2}·\left(2\right)\right)}^2$

$n^2+2n+{\left(1\right)}^2=13+{\left(1\right)}^2$

${\left(n+1\right)}^2=14$

$\sqrt{{\left(n+1\right)}^2}=\sqrt{14}$

$n+1=\pm \sqrt{14}$

$n+1-1=-1-\sqrt{14},n+1-1=-1+\sqrt{14}$

$n=-1-\sqrt{14},n=-1+\sqrt{14}$