-16t^2+54t+2.5=0

Solution for -16t^2+54t+2.5=0 equation:

-16t^2+54t+2.5=0

a = -16; b = 54; c = +2.5;
Δ = b2-4ac
Δ = 542-4·(-16)·2.5
Δ = 3076
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:

$\sqrt{\Delta}=\sqrt{3076}=\sqrt{4*769}=\sqrt{4}*\sqrt{769}=2\sqrt{769}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-2\sqrt{769}}{2*-16}=\frac{-54-2\sqrt{769}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+2\sqrt{769}}{2*-16}=\frac{-54+2\sqrt{769}}{-32}
$