2x^2-22x=-20

Solution for 2x^2-22x=-20 equation:

2x^2-22x=-20

We move all terms to the left:

2x^2-22x-(-20)=0

We add all the numbers together, and all the variables

2x^2-22x+20=0

a = 2; b = -22; c = +20;
Δ = b2-4ac
Δ = -222-4·2·20
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-18}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+18}{2*2}=\frac{40}{4}
=10
$