20+50t-16t^2=0

Solution for 20+50t-16t^2=0 equation:

20+50t-16t^2=0

a = -16; b = 50; c = +20;
Δ = b2-4ac
Δ = 502-4·(-16)·20
Δ = 3780
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:

$\sqrt{\Delta}=\sqrt{3780}=\sqrt{36*105}=\sqrt{36}*\sqrt{105}=6\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-6\sqrt{105}}{2*-16}=\frac{-50-6\sqrt{105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+6\sqrt{105}}{2*-16}=\frac{-50+6\sqrt{105}}{-32}
$