2x^2-38=4x

Solution for 2x^2-38=4x equation:

2x^2-38=4x

We move all terms to the left:

2x^2-38-(4x)=0

a = 2; b = -4; c = -38;
Δ = b2-4ac
Δ = -42-4·2·(-38)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{5}}{2*2}=\frac{4-8\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{5}}{2*2}=\frac{4+8\sqrt{5}}{4}
$