2x2-20x+40=0

Solution for 2x2-20x+40=0 equation:

2x^2-20x+40=0

a = 2; b = -20; c = +40;
Δ = b2-4ac
Δ = -202-4·2·40
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*2}=\frac{20-4\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*2}=\frac{20+4\sqrt{5}}{4}
$