-16t^2-15t+12=0

Solution for -16t^2-15t+12=0 equation:

-16t^2-15t+12=0

a = -16; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·(-16)·12
Δ = 993
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{993}}{2*-16}=\frac{15-\sqrt{993}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{993}}{2*-16}=\frac{15+\sqrt{993}}{-32}
$