1. Problem statement: $g(x)=\frac{-2x^2+7}{\sqrt{x^3+5x+6}-\sqrt{6}}$.\n2. Strategy: Rationalize the denominator by multiplying numerator and denominator by the conjugate $\sqrt{x^3+5x+6}+\sqrt{6}$.\n3. Multiply numerator and denominator: $g(x)=\frac{(-2x^2+7)(\sqrt{x^3+5x+6}+\sqrt{6})}{(\sqrt{x^3+5x+6}-\sqrt{6})(\sqrt{x^3+5x+6}+\sqrt{6})}$.\n4. Use the identity $ (a-b)(a+b)=a^2-b^2 $ to simplify the denominator: the denominator becomes $x^3+5x+6-6=x^3+5x$.\n5. Factor the denominator: $x^3+5x=x(x^2+5)$.\n6. Write the simplified expression: $g(x)=\frac{(-2x^2+7)(\sqrt{x^3+5x+6}+\sqrt{6})}{x(x^2+5)}$.\n7. Domain analysis: require $x^3+5x+6\ge 0$ for the square root to be real, and the original denominator must not be zero. Factor $x^3+5x+6=(x+1)(x^2-x+6)$ so the cubic is nonnegative for $x\ge -1$.\n8. Exclude points where the original denominator is zero: solve $\sqrt{x^3+5x+6}=\sqrt{6}$ which gives $x^3+5x=0$ and hence $x=0$.\n9. Final domain: $[-1,\infty)\setminus\{0\}$.\n10. Final answer: $g(x)=\frac{(-2x^2+7)(\sqrt{x^3+5x+6}+\sqrt{6})}{x(x^2+5)}$ valid for $x\in[-1,\infty)\setminus\{0\}$.\n