-16t^2+3t+9=0

Solution for -16t^2+3t+9=0 equation:

-16t^2+3t+9=0

a = -16; b = 3; c = +9;
Δ = b2-4ac
Δ = 32-4·(-16)·9
Δ = 585
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:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{65}}{2*-16}=\frac{-3-3\sqrt{65}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{65}}{2*-16}=\frac{-3+3\sqrt{65}}{-32}
$