2x^2-20x=-28

Solution for 2x^2-20x=-28 equation:

2x^2-20x=-28

We move all terms to the left:

2x^2-20x-(-28)=0

We add all the numbers together, and all the variables

2x^2-20x+28=0

a = 2; b = -20; c = +28;
Δ = b2-4ac
Δ = -202-4·2·28
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{11}}{2*2}=\frac{20-4\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{11}}{2*2}=\frac{20+4\sqrt{11}}{4}
$