0=-4.9t^2+35t+40

Solution for 0=-4.9t^2+35t+40 equation:

0=-4.9t^2+35t+40

We move all terms to the left:

0-(-4.9t^2+35t+40)=0

We add all the numbers together, and all the variables

-(-4.9t^2+35t+40)=0

We get rid of parentheses

4.9t^2-35t-40=0

a = 4.9; b = -35; c = -40;
Δ = b2-4ac
Δ = -352-4·4.9·(-40)
Δ = 2009
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:

$\sqrt{\Delta}=\sqrt{2009}=\sqrt{49*41}=\sqrt{49}*\sqrt{41}=7\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-7\sqrt{41}}{2*4.9}=\frac{35-7\sqrt{41}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+7\sqrt{41}}{2*4.9}=\frac{35+7\sqrt{41}}{9.8}
$