25c2+80c-64=0

Solution for 25c2+80c-64=0 equation:

25c^2+80c-64=0

a = 25; b = 80; c = -64;
Δ = b2-4ac
Δ = 802-4·25·(-64)
Δ = 12800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-80\sqrt{2}}{2*25}=\frac{-80-80\sqrt{2}}{50}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80\sqrt{2}}{2*25}=\frac{-80+80\sqrt{2}}{50}
$