-16t^2+50t+300=0

Solution for -16t^2+50t+300=0 equation:

-16t^2+50t+300=0

a = -16; b = 50; c = +300;
Δ = b2-4ac
Δ = 502-4·(-16)·300
Δ = 21700
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:

$\sqrt{\Delta}=\sqrt{21700}=\sqrt{100*217}=\sqrt{100}*\sqrt{217}=10\sqrt{217}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{217}}{2*-16}=\frac{-50-10\sqrt{217}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{217}}{2*-16}=\frac{-50+10\sqrt{217}}{-32}
$