-16t^2+20t+12=0

Solution for -16t^2+20t+12=0 equation:

-16t^2+20t+12=0

a = -16; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·(-16)·12
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{73}}{2*-16}=\frac{-20-4\sqrt{73}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{73}}{2*-16}=\frac{-20+4\sqrt{73}}{-32}
$