-15t^2+120t+10=0

Solution for -15t^2+120t+10=0 equation:

-15t^2+120t+10=0

a = -15; b = 120; c = +10;
Δ = b2-4ac
Δ = 1202-4·(-15)·10
Δ = 15000
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:

$\sqrt{\Delta}=\sqrt{15000}=\sqrt{2500*6}=\sqrt{2500}*\sqrt{6}=50\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-50\sqrt{6}}{2*-15}=\frac{-120-50\sqrt{6}}{-30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+50\sqrt{6}}{2*-15}=\frac{-120+50\sqrt{6}}{-30}
$