n2+20n+12=0

Solution for n2+20n+12=0 equation:

n2+20n+12=0

We add all the numbers together, and all the variables

n^2+20n+12=0

a = 1; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·1·12
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{22}}{2*1}=\frac{-20-4\sqrt{22}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{22}}{2*1}=\frac{-20+4\sqrt{22}}{2}
$