20=2t+6t^2

Solution for 20=2t+6t^2 equation:

20=2t+6t^2

We move all terms to the left:

20-(2t+6t^2)=0

We get rid of parentheses

-6t^2-2t+20=0

a = -6; b = -2; c = +20;
Δ = b2-4ac
Δ = -22-4·(-6)·20
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*-6}=\frac{-20}{-12}
=1+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-6}=\frac{24}{-12}
=-2
$