-16t^2+60t+50=0

Solution for -16t^2+60t+50=0 equation:

-16t^2+60t+50=0

a = -16; b = 60; c = +50;
Δ = b2-4ac
Δ = 602-4·(-16)·50
Δ = 6800
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:

$\sqrt{\Delta}=\sqrt{6800}=\sqrt{400*17}=\sqrt{400}*\sqrt{17}=20\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{17}}{2*-16}=\frac{-60-20\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{17}}{2*-16}=\frac{-60+20\sqrt{17}}{-32}
$