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

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

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

a = -16; b = -16; c = +9;
Δ = b2-4ac
Δ = -162-4·(-16)·9
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{13}}{2*-16}=\frac{16-8\sqrt{13}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{13}}{2*-16}=\frac{16+8\sqrt{13}}{-32}
$