2x^2-4x=39

Solution for 2x^2-4x=39 equation:

2x^2-4x=39

We move all terms to the left:

2x^2-4x-(39)=0

a = 2; b = -4; c = -39;
Δ = b2-4ac
Δ = -42-4·2·(-39)
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{82}}{2*2}=\frac{4-2\sqrt{82}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{82}}{2*2}=\frac{4+2\sqrt{82}}{4}
$