2x^2-32x-80=0

Solution for 2x^2-32x-80=0 equation:

2x^2-32x-80=0

a = 2; b = -32; c = -80;
Δ = b2-4ac
Δ = -322-4·2·(-80)
Δ = 1664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{26}}{2*2}=\frac{32-8\sqrt{26}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{26}}{2*2}=\frac{32+8\sqrt{26}}{4}
$