0=10-2t-4.9t^2

Solution for 0=10-2t-4.9t^2 equation:

0=10-2t-4.9t^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

4.9t^2+2t-10=0

a = 4.9; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·4.9·(-10)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10\sqrt{2}}{2*4.9}=\frac{-2-10\sqrt{2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10\sqrt{2}}{2*4.9}=\frac{-2+10\sqrt{2}}{9.8}
$