-16t^2-33t-16=0

Solution for -16t^2-33t-16=0 equation:

-16t^2-33t-16=0

a = -16; b = -33; c = -16;
Δ = b2-4ac
Δ = -332-4·(-16)·(-16)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-\sqrt{65}}{2*-16}=\frac{33-\sqrt{65}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{65}}{2*-16}=\frac{33+\sqrt{65}}{-32}
$