20=30t+5t^2

Solution for 20=30t+5t^2 equation:

20=30t+5t^2

We move all terms to the left:

20-(30t+5t^2)=0

We get rid of parentheses

-5t^2-30t+20=0

a = -5; b = -30; c = +20;
Δ = b2-4ac
Δ = -302-4·(-5)·20
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10\sqrt{13}}{2*-5}=\frac{30-10\sqrt{13}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10\sqrt{13}}{2*-5}=\frac{30+10\sqrt{13}}{-10}
$