-12t^2+25t+100=0

Solution for -12t^2+25t+100=0 equation:

-12t^2+25t+100=0

a = -12; b = 25; c = +100;
Δ = b2-4ac
Δ = 252-4·(-12)·100
Δ = 5425
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:

$\sqrt{\Delta}=\sqrt{5425}=\sqrt{25*217}=\sqrt{25}*\sqrt{217}=5\sqrt{217}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{217}}{2*-12}=\frac{-25-5\sqrt{217}}{-24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{217}}{2*-12}=\frac{-25+5\sqrt{217}}{-24}
$