-16t^2+12t+3=0

Solution for -16t^2+12t+3=0 equation:

-16t^2+12t+3=0

a = -16; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·(-16)·3
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{21}}{2*-16}=\frac{-12-4\sqrt{21}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{21}}{2*-16}=\frac{-12+4\sqrt{21}}{-32}
$