0=2x^2-20x+5

Solution for 0=2x^2-20x+5 equation:

0=2x^2-20x+5

We move all terms to the left:

0-(2x^2-20x+5)=0

We add all the numbers together, and all the variables

-(2x^2-20x+5)=0

We get rid of parentheses

-2x^2+20x-5=0

a = -2; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·(-2)·(-5)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{10}}{2*-2}=\frac{-20-6\sqrt{10}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{10}}{2*-2}=\frac{-20+6\sqrt{10}}{-4}
$