4x^2+12=64x

Solution for 4x^2+12=64x equation:

4x^2+12=64x

We move all terms to the left:

4x^2+12-(64x)=0

a = 4; b = -64; c = +12;
Δ = b2-4ac
Δ = -642-4·4·12
Δ = 3904
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:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{61}}{2*4}=\frac{64-8\sqrt{61}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{61}}{2*4}=\frac{64+8\sqrt{61}}{8}
$