2n^2+2n-20=0

Solution for 2n^2+2n-20=0 equation:

2n^2+2n-20=0

a = 2; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·2·(-20)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*2}=\frac{-2-2\sqrt{41}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*2}=\frac{-2+2\sqrt{41}}{4}
$