20t-16t^2+950=0

Solution for 20t-16t^2+950=0 equation:

20t-16t^2+950=0

a = -16; b = 20; c = +950;
Δ = b2-4ac
Δ = 202-4·(-16)·950
Δ = 61200
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:

$\sqrt{\Delta}=\sqrt{61200}=\sqrt{3600*17}=\sqrt{3600}*\sqrt{17}=60\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60\sqrt{17}}{2*-16}=\frac{-20-60\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60\sqrt{17}}{2*-16}=\frac{-20+60\sqrt{17}}{-32}
$