2x^2-38x+80=0

Solution for 2x^2-38x+80=0 equation:

2x^2-38x+80=0

a = 2; b = -38; c = +80;
Δ = b2-4ac
Δ = -382-4·2·80
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{201}}{2*2}=\frac{38-2\sqrt{201}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{201}}{2*2}=\frac{38+2\sqrt{201}}{4}
$