y = -3E-11x3 + 2E-07x2 + 3E-05x + 0.0077

I have the above equation and want to calculate x and so need to rearrange it:

x =

HELP!!

Why not use ^ to indicate powers? That is what SHIFT 6 is for.

I will use scientific notation instead of engineering notation:

y=-3\times 10^{-11}x^{3}+2\times 10^{-7}x^{2}+3\times 10^{-5}x+.0077

or in engineering notation, as you have:

y=-3E-11x^{3}+2E-7x^{2}+3E-5x+.0077

The two different methods are the same.

It is difficult to just solve for x. What is the y value? Do you have one? Are we to find the roots of the above cubic. That is, assuming y=0. That is the usual task.

Assuming y=0, then it has one real root and two complex roots.

The real one is 6.81884E3 or 6818.84

The two complex roots are -76.09+178.47i, \;\ -76.09-178.47i

Does this help?

Also, we can do this with standard notation:

y=\frac{-3}{100000000000}x^{3}+\frac{1}{5000000}x^{2}+\frac {3}{100000}x+\frac{77}{10000}

As you can see, just solving for x is a booger.:)

But, let me look at it and see what I can come up with. It may not be pleasant though.

Since the coefficient of x^3 is so small, we can probably disregard it and solve the resulting quadratic

That would probably be the best course of action. Is there more to the problem than what you have posted?

As a matter of fact, we can probably disregard the x^2 as well because it is small and solve the linear for x.

Doing that, we have y=\frac{3}{100000}x+\frac{77}{10000}

\fbox{x=\frac{100000}{3}y-\frac{770}{3}}

If we solve the quadratic for x, we get:

x=-75+5\sqrt{-1315+2\times 10^{5}y}, \;\ x=-75-5\sqrt{-1315+2\times 10^{5}y}

The cubic case is a monster. So, I would suggest the quadratic case since x^3 is so small.

You're welcome.