a^2-128a+256=0

Solution for a^2-128a+256=0 equation:

a^2-128a+256=0

a = 1; b = -128; c = +256;
Δ = b2-4ac
Δ = -1282-4·1·256
Δ = 15360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32\sqrt{15}}{2*1}=\frac{128-32\sqrt{15}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32\sqrt{15}}{2*1}=\frac{128+32\sqrt{15}}{2}
$