190-16t^2=0

Solution for 190-16t^2=0 equation:

190-16t^2=0

a = -16; b = 0; c = +190;
Δ = b2-4ac
Δ = 02-4·(-16)·190
Δ = 12160
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:

$\sqrt{\Delta}=\sqrt{12160}=\sqrt{64*190}=\sqrt{64}*\sqrt{190}=8\sqrt{190}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{190}}{2*-16}=\frac{0-8\sqrt{190}}{-32}
=-\frac{8\sqrt{190}}{-32}
=-\frac{\sqrt{190}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{190}}{2*-16}=\frac{0+8\sqrt{190}}{-32}
=\frac{8\sqrt{190}}{-32}
=\frac{\sqrt{190}}{-4}
$