n^2+11n=3n-4

Solution for n^2+11n=3n-4 equation:

n^2+11n=3n-4

We move all terms to the left:

n^2+11n-(3n-4)=0

We get rid of parentheses

n^2+11n-3n+4=0

We add all the numbers together, and all the variables

n^2+8n+4=0

a = 1; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·1·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*1}=\frac{-8-4\sqrt{3}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*1}=\frac{-8+4\sqrt{3}}{2}
$