0=2x^2-40x+20

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

0=2x^2-40x+20

We move all terms to the left:

0-(2x^2-40x+20)=0

We add all the numbers together, and all the variables

-(2x^2-40x+20)=0

We get rid of parentheses

-2x^2+40x-20=0

a = -2; b = 40; c = -20;
Δ = b2-4ac
Δ = 402-4·(-2)·(-20)
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-12\sqrt{10}}{2*-2}=\frac{-40-12\sqrt{10}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+12\sqrt{10}}{2*-2}=\frac{-40+12\sqrt{10}}{-4}
$