5n^2+22n=20

Solution for 5n^2+22n=20 equation:

5n^2+22n=20

We move all terms to the left:

5n^2+22n-(20)=0

a = 5; b = 22; c = -20;
Δ = b2-4ac
Δ = 222-4·5·(-20)
Δ = 884
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:

$\sqrt{\Delta}=\sqrt{884}=\sqrt{4*221}=\sqrt{4}*\sqrt{221}=2\sqrt{221}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{221}}{2*5}=\frac{-22-2\sqrt{221}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{221}}{2*5}=\frac{-22+2\sqrt{221}}{10}
$