1. **State the problem:** Solve the equation $$\sqrt{14x - 6} - 3 = x$$ for $x$. 2. **Isolate the square root:** Add 3 to both sides: $$\sqrt{14x - 6} = x + 3$$ 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{14x - 6}\right)^2 = (x + 3)^2$$ $$14x - 6 = (x + 3)^2$$ 4. **Expand the right side:** $$(x + 3)^2 = x^2 + 6x + 9$$ So, $$14x - 6 = x^2 + 6x + 9$$ 5. **Bring all terms to one side to form a quadratic equation:** $$0 = x^2 + 6x + 9 - 14x + 6$$ $$0 = x^2 - 8x + 15$$ 6. **Factor the quadratic:** $$x^2 - 8x + 15 = (x - 3)(x - 5)$$ 7. **Set each factor equal to zero and solve:** $$x - 3 = 0 \Rightarrow x = 3$$ $$x - 5 = 0 \Rightarrow x = 5$$ 8. **Check for extraneous solutions by substituting back into the original equation:** - For $x=3$: $$\sqrt{14(3) - 6} - 3 = \sqrt{42 - 6} - 3 = \sqrt{36} - 3 = 6 - 3 = 3$$ Right side is $x=3$, so $3=3$ is true. - For $x=5$: $$\sqrt{14(5) - 6} - 3 = \sqrt{70 - 6} - 3 = \sqrt{64} - 3 = 8 - 3 = 5$$ Right side is $x=5$, so $5=5$ is true. 9. **Both solutions are valid.** **Final answer:** $$x = 3, 5$$