0=-16t^2+45t+25

Solution for 0=-16t^2+45t+25 equation:

0=-16t^2+45t+25

We move all terms to the left:

0-(-16t^2+45t+25)=0

We add all the numbers together, and all the variables

-(-16t^2+45t+25)=0

We get rid of parentheses

16t^2-45t-25=0

a = 16; b = -45; c = -25;
Δ = b2-4ac
Δ = -452-4·16·(-25)
Δ = 3625
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:

$\sqrt{\Delta}=\sqrt{3625}=\sqrt{25*145}=\sqrt{25}*\sqrt{145}=5\sqrt{145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-5\sqrt{145}}{2*16}=\frac{45-5\sqrt{145}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+5\sqrt{145}}{2*16}=\frac{45+5\sqrt{145}}{32}
$