Solve quadratic equations by any of several algorithms. Factoring is usually the first choice, because it is often the easiest path. If factoring is not easy, there is a fail-safe path: the quadratic equation -- a little more computational intensive than factoring when it is easy, but the quadratic equation always produces an answer. The standard model for a quadratic is aX^2 + bX + c = 0, but this can be generalized to aX^2n + bX^n + c = 0 where n can be anything. This extends the power of quadratic solving tools to include many more trinomials.
Substitute negative exponents into the aX^2n + bX^n + c = 0 model with a substitution to get a quadratic, solve it and then restore the substitution. For example, the problem 9/X^3 = 8 + 1/X^6 does not look like a quadratic, but a little manipulation and a substitution allows a quadratic solution by factoring. 9/X^3 = 8 + 1/X^6 is equivalent to 1/X^6 -9/X^3 +8 = 0 is equivalent to X^-6 - 9X^-3 + 8 = 0. Substituting Y = X^-3 gives Y^2 - 9Y + 8 = 0 which can be easily factored to (Y - 1)(Y - 8) = 0 so Y = 1 and Y = 8 are both solutions. This means X^-3 = 1 and X^-3 = 8 or 1/X^3 = 1 and 1/X^3 = 8 so X = 1 and X = 1/2.
Use the substitution idea for fractions as well as for negative exponents. For example, consider the word problem" " X equals 8 times the square root of X minus 16, What is X?" This does not appear to be a quadratic equation but a little algebraic manipulation and a substitution make this a quadratic problem as well. X = 8X^1/2 - 16 so x - 8X^1/2 + 16 = 0. Substituting Y = X ^ 1/2 we have Y^2 - 8Y + 16 = 0 so (Y - 4)^2 = 0. This means Y = 4 so X^1/2 = 4 or X = 16.