250-16t^2=0

Solution for 250-16t^2=0 equation:

250-16t^2=0

a = -16; b = 0; c = +250;
Δ = b2-4ac
Δ = 02-4·(-16)·250
Δ = 16000
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:

$\sqrt{\Delta}=\sqrt{16000}=\sqrt{1600*10}=\sqrt{1600}*\sqrt{10}=40\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{10}}{2*-16}=\frac{0-40\sqrt{10}}{-32}
=-\frac{40\sqrt{10}}{-32}
=-\frac{5\sqrt{10}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{10}}{2*-16}=\frac{0+40\sqrt{10}}{-32}
=\frac{40\sqrt{10}}{-32}
=\frac{5\sqrt{10}}{-4}
$