20=20t+5t^2

Solution for 20=20t+5t^2 equation:

20=20t+5t^2

We move all terms to the left:

20-(20t+5t^2)=0

We get rid of parentheses

-5t^2-20t+20=0

a = -5; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·(-5)·20
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{2}}{2*-5}=\frac{20-20\sqrt{2}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{2}}{2*-5}=\frac{20+20\sqrt{2}}{-10}
$