-16t^2+5t+55=0

Solution for -16t^2+5t+55=0 equation:

-16t^2+5t+55=0

a = -16; b = 5; c = +55;
Δ = b2-4ac
Δ = 52-4·(-16)·55
Δ = 3545
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{-(5)-\sqrt{3545}}{2*-16}=\frac{-5-\sqrt{3545}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{3545}}{2*-16}=\frac{-5+\sqrt{3545}}{-32}
$