1. **State the problem:** Solve the equation $$\sqrt{7x - 10} = x$$ and check the solutions. 2. **Recall the domain:** The expression under the square root must be non-negative, so $$7x - 10 \geq 0 \Rightarrow x \geq \frac{10}{7} \approx 1.43$$. 3. **Square both sides:** To eliminate the square root, square both sides: $$\left(\sqrt{7x - 10}\right)^2 = x^2 \Rightarrow 7x - 10 = x^2$$ 4. **Rearrange into standard quadratic form:** $$x^2 - 7x + 10 = 0$$ 5. **Factor the quadratic:** $$x^2 - 7x + 10 = (x - 5)(x - 2) = 0$$ 6. **Find the roots:** $$x = 5 \quad \text{or} \quad x = 2$$ 7. **Check solutions in the original equation:** - For $$x=5$$: $$\sqrt{7(5) - 10} = \sqrt{35 - 10} = \sqrt{25} = 5$$ which equals $$x$$, so $$x=5$$ is valid. - For $$x=2$$: $$\sqrt{7(2) - 10} = \sqrt{14 - 10} = \sqrt{4} = 2$$ which equals $$x$$, so $$x=2$$ is valid. 8. **Check domain:** Both $$x=2$$ and $$x=5$$ satisfy $$x \geq \frac{10}{7}$$. **Final answer:** $$x = 2, 5$$