30=20t+4.9t^2

Solution for 30=20t+4.9t^2 equation:

30=20t+4.9t^2

We move all terms to the left:

30-(20t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-20t+30=0

a = -4.9; b = -20; c = +30;
Δ = b2-4ac
Δ = -202-4·(-4.9)·30
Δ = 988
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:

$\sqrt{\Delta}=\sqrt{988}=\sqrt{4*247}=\sqrt{4}*\sqrt{247}=2\sqrt{247}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{247}}{2*-4.9}=\frac{20-2\sqrt{247}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{247}}{2*-4.9}=\frac{20+2\sqrt{247}}{-9.8}
$