-16t^2+15t+5.5=0

Solution for -16t^2+15t+5.5=0 equation:

-16t^2+15t+5.5=0

a = -16; b = 15; c = +5.5;
Δ = b2-4ac
Δ = 152-4·(-16)·5.5
Δ = 577
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{-(15)-\sqrt{577}}{2*-16}=\frac{-15-\sqrt{577}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{577}}{2*-16}=\frac{-15+\sqrt{577}}{-32}
$