-16t^2+56t+33=0

Solution for -16t^2+56t+33=0 equation:

-16t^2+56t+33=0

a = -16; b = 56; c = +33;
Δ = b2-4ac
Δ = 562-4·(-16)·33
Δ = 5248
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:

$\sqrt{\Delta}=\sqrt{5248}=\sqrt{64*82}=\sqrt{64}*\sqrt{82}=8\sqrt{82}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-8\sqrt{82}}{2*-16}=\frac{-56-8\sqrt{82}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+8\sqrt{82}}{2*-16}=\frac{-56+8\sqrt{82}}{-32}
$