0=(16.0)t-4.9t^2

Solution for 0=(16.0)t-4.9t^2 equation:

0=(16.0)t-4.9t^2

We move all terms to the left:

0-((16.0)t-4.9t^2)=0

We add all the numbers together, and all the variables

-(16t-4.9t^2)+0=0

We add all the numbers together, and all the variables

-(16t-4.9t^2)=0

We get rid of parentheses

4.9t^2-16t=0

a = 4.9; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·4.9·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*4.9}=\frac{32}{9.8}
=3+2.6/9.8
$