100=20t+4.9t^2

Solution for 100=20t+4.9t^2 equation:

100=20t+4.9t^2

We move all terms to the left:

100-(20t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-20t+100=0

a = -4.9; b = -20; c = +100;
Δ = b2-4ac
Δ = -202-4·(-4.9)·100
Δ = 2360
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:

$\sqrt{\Delta}=\sqrt{2360}=\sqrt{4*590}=\sqrt{4}*\sqrt{590}=2\sqrt{590}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{590}}{2*-4.9}=\frac{20-2\sqrt{590}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{590}}{2*-4.9}=\frac{20+2\sqrt{590}}{-9.8}
$