25x^2-160x+64=0

Solution for 25x^2-160x+64=0 equation:

25x^2-160x+64=0

a = 25; b = -160; c = +64;
Δ = b2-4ac
Δ = -1602-4·25·64
Δ = 19200
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:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-80\sqrt{3}}{2*25}=\frac{160-80\sqrt{3}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+80\sqrt{3}}{2*25}=\frac{160+80\sqrt{3}}{50}
$