0=-16t^2+2.5

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

0=-16t^2+2.5

We move all terms to the left:

0-(-16t^2+2.5)=0

We add all the numbers together, and all the variables

-(-16t^2+2.5)=0

We get rid of parentheses

16t^2-2.5=0

a = 16; b = 0; c = -2.5;
Δ = b2-4ac
Δ = 02-4·16·(-2.5)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*16}=\frac{0-4\sqrt{10}}{32}
=-\frac{4\sqrt{10}}{32}
=-\frac{\sqrt{10}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*16}=\frac{0+4\sqrt{10}}{32}
=\frac{4\sqrt{10}}{32}
=\frac{\sqrt{10}}{8}
$