2x^2+36x+80=0

Solution for 2x^2+36x+80=0 equation:

2x^2+36x+80=0

a = 2; b = 36; c = +80;
Δ = b2-4ac
Δ = 362-4·2·80
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{41}}{2*2}=\frac{-36-4\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{41}}{2*2}=\frac{-36+4\sqrt{41}}{4}
$