25x^2-96x+68=0

Solution for 25x^2-96x+68=0 equation:

25x^2-96x+68=0

a = 25; b = -96; c = +68;
Δ = b2-4ac
Δ = -962-4·25·68
Δ = 2416
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:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-4\sqrt{151}}{2*25}=\frac{96-4\sqrt{151}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+4\sqrt{151}}{2*25}=\frac{96+4\sqrt{151}}{50}
$