Answer
$$\frac{x^2+3x}{2x^2-8x-10}$$
Work Step by Step
We first recall that dividing is the same as multiplying by the inverse. Using the rules of exponents and cancelling common factors in the numerator and the denominator, we find:
$$\frac{\left(x^2-6x-27\right)}{2x^2+2x}\times \frac{x^2}{\left(x^2-14x+45\right)}\\ \frac{\left(x^2-6x-27\right)x^2}{\left(2x^2+2x\right)\left(x^2-14x+45\right)}\\ \frac{x\left(x+3\right)\left(x-9\right)}{2\left(x+1\right)\left(x-5\right)\left(x-9\right)}\\ \frac{x^2+3x}{2x^2-8x-10}$$
