-30t^2+48t+36=0

Solution for -30t^2+48t+36=0 equation:

-30t^2+48t+36=0

a = -30; b = 48; c = +36;
Δ = b2-4ac
Δ = 482-4·(-30)·36
Δ = 6624
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:

$\sqrt{\Delta}=\sqrt{6624}=\sqrt{144*46}=\sqrt{144}*\sqrt{46}=12\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{46}}{2*-30}=\frac{-48-12\sqrt{46}}{-60}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{46}}{2*-30}=\frac{-48+12\sqrt{46}}{-60}
$