25x^2+60x-16=0

Solution for 25x^2+60x-16=0 equation:

25x^2+60x-16=0

a = 25; b = 60; c = -16;
Δ = b2-4ac
Δ = 602-4·25·(-16)
Δ = 5200
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:

$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{13}}{2*25}=\frac{-60-20\sqrt{13}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{13}}{2*25}=\frac{-60+20\sqrt{13}}{50}
$