-25t^2+100t+75=0

Solution for -25t^2+100t+75=0 equation:

-25t^2+100t+75=0

a = -25; b = 100; c = +75;
Δ = b2-4ac
Δ = 1002-4·(-25)·75
Δ = 17500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{17500}=\sqrt{2500*7}=\sqrt{2500}*\sqrt{7}=50\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-50\sqrt{7}}{2*-25}=\frac{-100-50\sqrt{7}}{-50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+50\sqrt{7}}{2*-25}=\frac{-100+50\sqrt{7}}{-50}
$