1. **State the problem:** Solve the equation $$\sqrt{-2x - 5} - 4 = x$$ for $x$. 2. **Isolate the square root:** Add 4 to both sides: $$\sqrt{-2x - 5} = x + 4$$ 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{-2x - 5}\right)^2 = (x + 4)^2$$ $$-2x - 5 = (x + 4)^2$$ 4. **Expand the right side:** $$(x + 4)^2 = x^2 + 8x + 16$$ So, $$-2x - 5 = x^2 + 8x + 16$$ 5. **Bring all terms to one side to form a quadratic equation:** $$0 = x^2 + 8x + 16 + 2x + 5$$ $$0 = x^2 + 10x + 21$$ 6. **Factor the quadratic:** $$x^2 + 10x + 21 = (x + 3)(x + 7)$$ 7. **Set each factor equal to zero:** $$x + 3 = 0 \Rightarrow x = -3$$ $$x + 7 = 0 \Rightarrow x = -7$$ 8. **Check for extraneous solutions by substituting back into the original equation:** - For $x = -3$: $$\sqrt{-2(-3) - 5} - 4 = \sqrt{6 - 5} - 4 = \sqrt{1} - 4 = 1 - 4 = -3$$ Right side is $x = -3$, so it satisfies the equation. - For $x = -7$: $$\sqrt{-2(-7) - 5} - 4 = \sqrt{14 - 5} - 4 = \sqrt{9} - 4 = 3 - 4 = -1$$ Right side is $x = -7$, which is not equal to -1, so $x = -7$ is extraneous. **Final answer:** $x = -3$