0=n^2+2n-11

Solution for 0=n^2+2n-11 equation:

0=n^2+2n-11

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-n^2-2n+11=0

We add all the numbers together, and all the variables

-1n^2-2n+11=0

a = -1; b = -2; c = +11;
Δ = b2-4ac
Δ = -22-4·(-1)·11
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{-(-2)-4\sqrt{3}}{2*-1}=\frac{2-4\sqrt{3}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{3}}{2*-1}=\frac{2+4\sqrt{3}}{-2}
$