-16t^2+5t+50=0

Solution for -16t^2+5t+50=0 equation:

-16t^2+5t+50=0

a = -16; b = 5; c = +50;
Δ = b2-4ac
Δ = 52-4·(-16)·50
Δ = 3225
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:

$\sqrt{\Delta}=\sqrt{3225}=\sqrt{25*129}=\sqrt{25}*\sqrt{129}=5\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{129}}{2*-16}=\frac{-5-5\sqrt{129}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{129}}{2*-16}=\frac{-5+5\sqrt{129}}{-32}
$