Quadratic equations are of the form:

ax² + bx + c = y, a, b, and c are just the constants before the variables.

Like a would be the constant term before the x² and so forth.

Your equation above does not describe something falling however.

1) There is no initial height.

2) The initial velocity would suggest that it was going up and not down (though you can have positive values for problems like this)

3) If we are using gravity, then the acceleration would be negative and not positive.

A free-fall position formula should take the form:

Yf = Yi + vt - .5gt²; Yf = final height, Yi = initial height, v = initial velocity

The way that you have it written states that:

Yi = 40m, v = +10m/s, a = -9.8m/s²

So,

0 = 40 + 10t - 4.9t² (when it hits the ground Yf = 0)

In normal quadratic form:

-4.9t² + 10t + 40 = 0

Using the quadradic equation:

[-10 ± √(100 + 784)] / -9.8

t = -2.01348... and 4.054299...

Since time cannot be negative;

t ≈ 4.0543s

Sorry Mathsyperson, you forgot to use the correct conditions stipulated by the problem. Don't let Mathsy's error confuse you about his math ability, he is very astute.