0=-4.9t^2+3.5t+10

Solution for 0=-4.9t^2+3.5t+10 equation:

0=-4.9t^2+3.5t+10

We move all terms to the left:

0-(-4.9t^2+3.5t+10)=0

We add all the numbers together, and all the variables

-(-4.9t^2+3.5t+10)=0

We get rid of parentheses

4.9t^2-3.5t-10=0

a = 4.9; b = -3.5; c = -10;
Δ = b2-4ac
Δ = -3.52-4·4.9·(-10)
Δ = 208.25
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{-(-3.5)-\sqrt{208.25}}{2*4.9}=\frac{3.5-\sqrt{208.25}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.5)+\sqrt{208.25}}{2*4.9}=\frac{3.5+\sqrt{208.25}}{9.8}
$