2x^2-8x=28

Solution for 2x^2-8x=28 equation:

2x^2-8x=28

We move all terms to the left:

2x^2-8x-(28)=0

a = 2; b = -8; c = -28;
Δ = b2-4ac
Δ = -82-4·2·(-28)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12\sqrt{2}}{2*2}=\frac{8-12\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12\sqrt{2}}{2*2}=\frac{8+12\sqrt{2}}{4}
$