50-16t^2=0

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

50-16t^2=0

a = -16; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-16)·50
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*-16}=\frac{0-40\sqrt{2}}{-32}
=-\frac{40\sqrt{2}}{-32}
=-\frac{5\sqrt{2}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*-16}=\frac{0+40\sqrt{2}}{-32}
=\frac{40\sqrt{2}}{-32}
=\frac{5\sqrt{2}}{-4}
$