-16t^2+95=0

Solution for -16t^2+95=0 equation:

-16t^2+95=0

a = -16; b = 0; c = +95;
Δ = b2-4ac
Δ = 02-4·(-16)·95
Δ = 6080
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:

$\sqrt{\Delta}=\sqrt{6080}=\sqrt{64*95}=\sqrt{64}*\sqrt{95}=8\sqrt{95}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{95}}{2*-16}=\frac{0-8\sqrt{95}}{-32}
=-\frac{8\sqrt{95}}{-32}
=-\frac{\sqrt{95}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{95}}{2*-16}=\frac{0+8\sqrt{95}}{-32}
=\frac{8\sqrt{95}}{-32}
=\frac{\sqrt{95}}{-4}
$