4x^2-128x+160=0

Solution for 4x^2-128x+160=0 equation:

4x^2-128x+160=0

a = 4; b = -128; c = +160;
Δ = b2-4ac
Δ = -1282-4·4·160
Δ = 13824
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:

$\sqrt{\Delta}=\sqrt{13824}=\sqrt{2304*6}=\sqrt{2304}*\sqrt{6}=48\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-48\sqrt{6}}{2*4}=\frac{128-48\sqrt{6}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+48\sqrt{6}}{2*4}=\frac{128+48\sqrt{6}}{8}
$