-12t^2-25t+500=0

Solution for -12t^2-25t+500=0 equation:

-12t^2-25t+500=0

a = -12; b = -25; c = +500;
Δ = b2-4ac
Δ = -252-4·(-12)·500
Δ = 24625
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:

$\sqrt{\Delta}=\sqrt{24625}=\sqrt{25*985}=\sqrt{25}*\sqrt{985}=5\sqrt{985}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{985}}{2*-12}=\frac{25-5\sqrt{985}}{-24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{985}}{2*-12}=\frac{25+5\sqrt{985}}{-24}
$