t=-16t^2+32t+20

Solution for t=-16t^2+32t+20 equation:

t=-16t^2+32t+20

We move all terms to the left:

t-(-16t^2+32t+20)=0

We get rid of parentheses

16t^2-32t+t-20=0

We add all the numbers together, and all the variables

16t^2-31t-20=0

a = 16; b = -31; c = -20;
Δ = b2-4ac
Δ = -312-4·16·(-20)
Δ = 2241
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:

$\sqrt{\Delta}=\sqrt{2241}=\sqrt{9*249}=\sqrt{9}*\sqrt{249}=3\sqrt{249}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-3\sqrt{249}}{2*16}=\frac{31-3\sqrt{249}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+3\sqrt{249}}{2*16}=\frac{31+3\sqrt{249}}{32}
$