10=10+2t+-4.9t^2

Solution for 10=10+2t+-4.9t^2 equation:

10=10+2t+-4.9t^2

We move all terms to the left:

10-(10+2t+-4.9t^2)=0

We use the square of the difference formula

-(10+2t-4.9t^2)+10=0

We get rid of parentheses

4.9t^2-2t-10+10=0

We add all the numbers together, and all the variables

4.9t^2-2t=0

a = 4.9; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·4.9·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*4.9}=\frac{4}{9.8}
=4/9.8
$