2n^2-20n-32=0

Solution for 2n^2-20n-32=0 equation:

2n^2-20n-32=0

a = 2; b = -20; c = -32;
Δ = b2-4ac
Δ = -202-4·2·(-32)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{41}}{2*2}=\frac{20-4\sqrt{41}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{41}}{2*2}=\frac{20+4\sqrt{41}}{4}
$