-16t^2+6t+192=0

Solution for -16t^2+6t+192=0 equation:

-16t^2+6t+192=0

a = -16; b = 6; c = +192;
Δ = b2-4ac
Δ = 62-4·(-16)·192
Δ = 12324
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:

$\sqrt{\Delta}=\sqrt{12324}=\sqrt{4*3081}=\sqrt{4}*\sqrt{3081}=2\sqrt{3081}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3081}}{2*-16}=\frac{-6-2\sqrt{3081}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3081}}{2*-16}=\frac{-6+2\sqrt{3081}}{-32}
$