0=4.9t^2-3.59t+.125

Solution for 0=4.9t^2-3.59t+.125 equation:

0=4.9t^2-3.59t+.125

We move all terms to the left:

0-(4.9t^2-3.59t+.125)=0

We add all the numbers together, and all the variables

-(4.9t^2-3.59t+.125)=0

We get rid of parentheses

-4.9t^2+3.59t-.125=0

We add all the numbers together, and all the variables

-4.9t^2+3.59t-0.125=0

a = -4.9; b = 3.59; c = -0.125;
Δ = b2-4ac
Δ = 3.592-4·(-4.9)·(-0.125)
Δ = 10.4381
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3.59)-\sqrt{10.4381}}{2*-4.9}=\frac{-3.59-\sqrt{10.4381}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.59)+\sqrt{10.4381}}{2*-4.9}=\frac{-3.59+\sqrt{10.4381}}{-9.8}
$