-16t^2+12t+2.5=0

Solution for -16t^2+12t+2.5=0 equation:

-16t^2+12t+2.5=0

a = -16; b = 12; c = +2.5;
Δ = b2-4ac
Δ = 122-4·(-16)·2.5
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{19}}{2*-16}=\frac{-12-4\sqrt{19}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{19}}{2*-16}=\frac{-12+4\sqrt{19}}{-32}
$