20(t)=-16t^2+25

Solution for 20(t)=-16t^2+25 equation:

20(t)=-16t^2+25

We move all terms to the left:

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

We get rid of parentheses

16t^2+20t-25=0

a = 16; b = 20; c = -25;
Δ = b2-4ac
Δ = 202-4·16·(-25)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{5}}{2*16}=\frac{-20-20\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{5}}{2*16}=\frac{-20+20\sqrt{5}}{32}
$