4x^2+8=16x

Solution for 4x^2+8=16x equation:

4x^2+8=16x

We move all terms to the left:

4x^2+8-(16x)=0

a = 4; b = -16; c = +8;
Δ = b2-4ac
Δ = -162-4·4·8
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{2}}{2*4}=\frac{16-8\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{2}}{2*4}=\frac{16+8\sqrt{2}}{8}
$