-16t^2+36t+9=0

Solution for -16t^2+36t+9=0 equation:

-16t^2+36t+9=0

a = -16; b = 36; c = +9;
Δ = b2-4ac
Δ = 362-4·(-16)·9
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12\sqrt{13}}{2*-16}=\frac{-36-12\sqrt{13}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12\sqrt{13}}{2*-16}=\frac{-36+12\sqrt{13}}{-32}
$