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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-19t-2=0

a = 16; b = -19; c = -2;
Δ = b2-4ac
Δ = -192-4·16·(-2)
Δ = 489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{489}}{2*16}=\frac{19-\sqrt{489}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{489}}{2*16}=\frac{19+\sqrt{489}}{32}
$