0=8t+-4.9t^2

Solution for 0=8t+-4.9t^2 equation:

0=8t+-4.9t^2

We move all terms to the left:

0-(8t+-4.9t^2)=0

We add all the numbers together, and all the variables

-(8t+-4.9t^2)=0

We use the square of the difference formula

-(8t-4.9t^2)=0

We get rid of parentheses

4.9t^2-8t=0

a = 4.9; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·4.9·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*4.9}=\frac{16}{9.8}
=1+1.9375/3.0625
$