-16t^2+45t+220=0

Solution for -16t^2+45t+220=0 equation:

-16t^2+45t+220=0

a = -16; b = 45; c = +220;
Δ = b2-4ac
Δ = 452-4·(-16)·220
Δ = 16105
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{-(45)-\sqrt{16105}}{2*-16}=\frac{-45-\sqrt{16105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{16105}}{2*-16}=\frac{-45+\sqrt{16105}}{-32}
$