4x^2-20=4x

Solution for 4x^2-20=4x equation:

4x^2-20=4x

We move all terms to the left:

4x^2-20-(4x)=0

a = 4; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·4·(-20)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{21}}{2*4}=\frac{4-4\sqrt{21}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{21}}{2*4}=\frac{4+4\sqrt{21}}{8}
$