4t+9.8*t^2=20

Solution for 4t+9.8*t^2=20 equation:

4t+9.8*t^2=20

We move all terms to the left:

4t+9.8*t^2-(20)=0

We add all the numbers together, and all the variables

9.8t^2+4t-20=0

a = 9.8; b = 4; c = -20;
Δ = b2-4ac
Δ = 42-4·9.8·(-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{-(4)-20\sqrt{2}}{2*9.8}=\frac{-4-20\sqrt{2}}{19.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20\sqrt{2}}{2*9.8}=\frac{-4+20\sqrt{2}}{19.6}
$