20=2+35t-16t^2

Solution for 20=2+35t-16t^2 equation:

20=2+35t-16t^2

We move all terms to the left:

20-(2+35t-16t^2)=0

We get rid of parentheses

16t^2-35t-2+20=0

We add all the numbers together, and all the variables

16t^2-35t+18=0

a = 16; b = -35; c = +18;
Δ = b2-4ac
Δ = -352-4·16·18
Δ = 73
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{-(-35)-\sqrt{73}}{2*16}=\frac{35-\sqrt{73}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{73}}{2*16}=\frac{35+\sqrt{73}}{32}
$