0=-4.9t^2-3.9t+10

Solution for 0=-4.9t^2-3.9t+10 equation:

0=-4.9t^2-3.9t+10

We move all terms to the left:

0-(-4.9t^2-3.9t+10)=0

We add all the numbers together, and all the variables

-(-4.9t^2-3.9t+10)=0

We get rid of parentheses

4.9t^2+3.9t-10=0

a = 4.9; b = 3.9; c = -10;
Δ = b2-4ac
Δ = 3.92-4·4.9·(-10)
Δ = 211.21
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{-(3.9)-\sqrt{211.21}}{2*4.9}=\frac{-3.9-\sqrt{211.21}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.9)+\sqrt{211.21}}{2*4.9}=\frac{-3.9+\sqrt{211.21}}{9.8}
$