2x^2-48x+256=0

Solution for 2x^2-48x+256=0 equation:

2x^2-48x+256=0

a = 2; b = -48; c = +256;
Δ = b2-4ac
Δ = -482-4·2·256
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*2}=\frac{32}{4}
=8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*2}=\frac{64}{4}
=16
$