3n^2+22n+20=0

Solution for 3n^2+22n+20=0 equation:

3n^2+22n+20=0

a = 3; b = 22; c = +20;
Δ = b2-4ac
Δ = 222-4·3·20
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{61}}{2*3}=\frac{-22-2\sqrt{61}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{61}}{2*3}=\frac{-22+2\sqrt{61}}{6}
$