1. **State the problem:** Solve the equation $\sqrt{2x + 6} + 4 = x + 3$ for $x$. 2. **Isolate the square root:** Subtract 4 from both sides: $$\sqrt{2x + 6} = x + 3 - 4$$ $$\sqrt{2x + 6} = x - 1$$ 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{2x + 6}\right)^2 = (x - 1)^2$$ $$2x + 6 = (x - 1)^2$$ 4. **Expand the right side:** $$(x - 1)^2 = x^2 - 2x + 1$$ So, $$2x + 6 = x^2 - 2x + 1$$ 5. **Bring all terms to one side to form a quadratic equation:** $$0 = x^2 - 2x + 1 - 2x - 6$$ $$0 = x^2 - 4x - 5$$ 6. **Factor the quadratic:** $$x^2 - 4x - 5 = (x - 5)(x + 1)$$ 7. **Solve for $x$:** $$x - 5 = 0 \Rightarrow x = 5$$ $$x + 1 = 0 \Rightarrow x = -1$$ 8. **Check for extraneous solutions by substituting back into the original equation:** - For $x=5$: $$\sqrt{2(5) + 6} + 4 = \sqrt{10 + 6} + 4 = \sqrt{16} + 4 = 4 + 4 = 8$$ Right side: $$5 + 3 = 8$$ Valid. - For $x=-1$: $$\sqrt{2(-1) + 6} + 4 = \sqrt{-2 + 6} + 4 = \sqrt{4} + 4 = 2 + 4 = 6$$ Right side: $$-1 + 3 = 2$$ Not valid. **Final solution set:** $$\boxed{\{5\}}$$