16t+-4.9t^2=0

Solution for 16t+-4.9t^2=0 equation:

16t+-4.9t^2=0

We add all the numbers together, and all the variables

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