n^2+8n=11

Solution for n^2+8n=11 equation:

n^2+8n=11

We move all terms to the left:

n^2+8n-(11)=0

a = 1; b = 8; c = -11;
Δ = b2-4ac
Δ = 82-4·1·(-11)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{3}}{2*1}=\frac{-8-6\sqrt{3}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{3}}{2*1}=\frac{-8+6\sqrt{3}}{2}
$