4x^2-136x+256=0

Solution for 4x^2-136x+256=0 equation:

4x^2-136x+256=0

a = 4; b = -136; c = +256;
Δ = b2-4ac
Δ = -1362-4·4·256
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-136)-120}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-136)+120}{2*4}=\frac{256}{8}
=32
$