25x^2-96x+64=0

Solution for 25x^2-96x+64=0 equation:

25x^2-96x+64=0

a = 25; b = -96; c = +64;
Δ = b2-4ac
Δ = -962-4·25·64
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16\sqrt{11}}{2*25}=\frac{96-16\sqrt{11}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16\sqrt{11}}{2*25}=\frac{96+16\sqrt{11}}{50}
$