-16t^2+33t+8=0

Solution for -16t^2+33t+8=0 equation:

-16t^2+33t+8=0

a = -16; b = 33; c = +8;
Δ = b2-4ac
Δ = 332-4·(-16)·8
Δ = 1601
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{1601}}{2*-16}=\frac{-33-\sqrt{1601}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{1601}}{2*-16}=\frac{-33+\sqrt{1601}}{-32}
$