100=(4.9)(t*t)+2t

Solution for 100=(4.9)(t*t)+2t equation:

100=(4.9)(t*t)+2t

We move all terms to the left:

100-((4.9)(t*t)+2t)=0

We add all the numbers together, and all the variables

-((4.9)(+t*t)+2t)+100=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+4.9t^2)+2t), so:

(+4.9t^2)+2t

We get rid of parentheses

4.9t^2+2t

Back to the equation:

-(4.9t^2+2t)


We get rid of parentheses

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

a = -4.9; b = -2; c = +100;
Δ = b2-4ac
Δ = -22-4·(-4.9)·100
Δ = 1964
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:

$\sqrt{\Delta}=\sqrt{1964}=\sqrt{4*491}=\sqrt{4}*\sqrt{491}=2\sqrt{491}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{491}}{2*-4.9}=\frac{2-2\sqrt{491}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{491}}{2*-4.9}=\frac{2+2\sqrt{491}}{-9.8}
$