41=2x^2-4x

Solution for 41=2x^2-4x equation:

41=2x^2-4x

We move all terms to the left:

41-(2x^2-4x)=0

We get rid of parentheses

-2x^2+4x+41=0

a = -2; b = 4; c = +41;
Δ = b2-4ac
Δ = 42-4·(-2)·41
Δ = 344
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:

$\sqrt{\Delta}=\sqrt{344}=\sqrt{4*86}=\sqrt{4}*\sqrt{86}=2\sqrt{86}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{86}}{2*-2}=\frac{-4-2\sqrt{86}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{86}}{2*-2}=\frac{-4+2\sqrt{86}}{-4}
$