100=4.9t^2+5t

Solution for 100=4.9t^2+5t equation:

100=4.9t^2+5t

We move all terms to the left:

100-(4.9t^2+5t)=0

We get rid of parentheses

-4.9t^2-5t+100=0

a = -4.9; b = -5; c = +100;
Δ = b2-4ac
Δ = -52-4·(-4.9)·100
Δ = 1985
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:

$\sqrt{\Delta}=\sqrt{1985}=\sqrt{1*1985}=\sqrt{1}*\sqrt{1985}=1\sqrt{1985}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1\sqrt{1985}}{2*-4.9}=\frac{5-1\sqrt{1985}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1\sqrt{1985}}{2*-4.9}=\frac{5+1\sqrt{1985}}{-9.8}
$