-16t^2+96t-3=0

Solution for -16t^2+96t-3=0 equation:

-16t^2+96t-3=0

a = -16; b = 96; c = -3;
Δ = b2-4ac
Δ = 962-4·(-16)·(-3)
Δ = 9024
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:

$\sqrt{\Delta}=\sqrt{9024}=\sqrt{64*141}=\sqrt{64}*\sqrt{141}=8\sqrt{141}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-8\sqrt{141}}{2*-16}=\frac{-96-8\sqrt{141}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+8\sqrt{141}}{2*-16}=\frac{-96+8\sqrt{141}}{-32}
$