0=4.9t^2+10t+2

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

0=4.9t^2+10t+2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -4.9; b = -10; c = -2;
Δ = b2-4ac
Δ = -102-4·(-4.9)·(-2)
Δ = 60.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-\sqrt{60.8}}{2*-4.9}=\frac{10-\sqrt{60.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{60.8}}{2*-4.9}=\frac{10+\sqrt{60.8}}{-9.8}
$