-3t^2+96t+40=0

Solution for -3t^2+96t+40=0 equation:

-3t^2+96t+40=0

a = -3; b = 96; c = +40;
Δ = b2-4ac
Δ = 962-4·(-3)·40
Δ = 9696
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:

$\sqrt{\Delta}=\sqrt{9696}=\sqrt{16*606}=\sqrt{16}*\sqrt{606}=4\sqrt{606}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-4\sqrt{606}}{2*-3}=\frac{-96-4\sqrt{606}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+4\sqrt{606}}{2*-3}=\frac{-96+4\sqrt{606}}{-6}
$