2x^2+80x-408=0

Solution for 2x^2+80x-408=0 equation:

2x^2+80x-408=0

a = 2; b = 80; c = -408;
Δ = b2-4ac
Δ = 802-4·2·(-408)
Δ = 9664
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:

$\sqrt{\Delta}=\sqrt{9664}=\sqrt{64*151}=\sqrt{64}*\sqrt{151}=8\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{151}}{2*2}=\frac{-80-8\sqrt{151}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{151}}{2*2}=\frac{-80+8\sqrt{151}}{4}
$