2x^2-20x=68

Solution for 2x^2-20x=68 equation:

2x^2-20x=68

We move all terms to the left:

2x^2-20x-(68)=0

a = 2; b = -20; c = -68;
Δ = b2-4ac
Δ = -202-4·2·(-68)
Δ = 944
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:

$\sqrt{\Delta}=\sqrt{944}=\sqrt{16*59}=\sqrt{16}*\sqrt{59}=4\sqrt{59}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{59}}{2*2}=\frac{20-4\sqrt{59}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{59}}{2*2}=\frac{20+4\sqrt{59}}{4}
$