0=-16t^2+20t+45

Solution for 0=-16t^2+20t+45 equation:

0=-16t^2+20t+45

We move all terms to the left:

0-(-16t^2+20t+45)=0

We add all the numbers together, and all the variables

-(-16t^2+20t+45)=0

We get rid of parentheses

16t^2-20t-45=0

a = 16; b = -20; c = -45;
Δ = b2-4ac
Δ = -202-4·16·(-45)
Δ = 3280
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:

$\sqrt{\Delta}=\sqrt{3280}=\sqrt{16*205}=\sqrt{16}*\sqrt{205}=4\sqrt{205}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{205}}{2*16}=\frac{20-4\sqrt{205}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{205}}{2*16}=\frac{20+4\sqrt{205}}{32}
$