-16t^2+45t+50=0

Solution for -16t^2+45t+50=0 equation:

-16t^2+45t+50=0

a = -16; b = 45; c = +50;
Δ = b2-4ac
Δ = 452-4·(-16)·50
Δ = 5225
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:

$\sqrt{\Delta}=\sqrt{5225}=\sqrt{25*209}=\sqrt{25}*\sqrt{209}=5\sqrt{209}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-5\sqrt{209}}{2*-16}=\frac{-45-5\sqrt{209}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+5\sqrt{209}}{2*-16}=\frac{-45+5\sqrt{209}}{-32}
$