0=4t-4.9t^2+40

Solution for 0=4t-4.9t^2+40 equation:

0=4t-4.9t^2+40

We move all terms to the left:

0-(4t-4.9t^2+40)=0

We add all the numbers together, and all the variables

-(4t-4.9t^2+40)=0

We get rid of parentheses

4.9t^2-4t-40=0

a = 4.9; b = -4; c = -40;
Δ = b2-4ac
Δ = -42-4·4.9·(-40)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20\sqrt{2}}{2*4.9}=\frac{4-20\sqrt{2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20\sqrt{2}}{2*4.9}=\frac{4+20\sqrt{2}}{9.8}
$