20=2t+3t^2

Solution for 20=2t+3t^2 equation:

20=2t+3t^2

We move all terms to the left:

20-(2t+3t^2)=0

We get rid of parentheses

-3t^2-2t+20=0

a = -3; b = -2; c = +20;
Δ = b2-4ac
Δ = -22-4·(-3)·20
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{61}}{2*-3}=\frac{2-2\sqrt{61}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{61}}{2*-3}=\frac{2+2\sqrt{61}}{-6}
$