64x^2-320x+25=0

Solution for 64x^2-320x+25=0 equation:

64x^2-320x+25=0

a = 64; b = -320; c = +25;
Δ = b2-4ac
Δ = -3202-4·64·25
Δ = 96000
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:

$\sqrt{\Delta}=\sqrt{96000}=\sqrt{6400*15}=\sqrt{6400}*\sqrt{15}=80\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-320)-80\sqrt{15}}{2*64}=\frac{320-80\sqrt{15}}{128}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-320)+80\sqrt{15}}{2*64}=\frac{320+80\sqrt{15}}{128}
$