6+96t-16t2=0

Solution for 6+96t-16t2=0 equation:

6+96t-16t^2=0

a = -16; b = 96; c = +6;
Δ = b2-4ac
Δ = 962-4·(-16)·6
Δ = 9600
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:

$\sqrt{\Delta}=\sqrt{9600}=\sqrt{1600*6}=\sqrt{1600}*\sqrt{6}=40\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-40\sqrt{6}}{2*-16}=\frac{-96-40\sqrt{6}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+40\sqrt{6}}{2*-16}=\frac{-96+40\sqrt{6}}{-32}
$