n^2+2n+20=22

Solution for n^2+2n+20=22 equation:

n^2+2n+20=22

We move all terms to the left:

n^2+2n+20-(22)=0

We add all the numbers together, and all the variables

n^2+2n-2=0

a = 1; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·1·(-2)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*1}=\frac{-2-2\sqrt{3}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*1}=\frac{-2+2\sqrt{3}}{2}
$