1. **State the problem:** Simplify the expression $\sqrt{9x^2+6x-3}$. 2. **Identify the expression inside the square root:** The radicand is $9x^2 + 6x - 3$. 3. **Try to factor the quadratic inside the square root:** We look for factors of $9x^2 + 6x - 3$. 4. **Factor out the greatest common factor (GCF):** $$\sqrt{9x^2 + 6x - 3} = \sqrt{3(3x^2 + 2x - 1)}$$ 5. **Factor the quadratic $3x^2 + 2x - 1$:** We look for two numbers that multiply to $3 \times (-1) = -3$ and add to $2$. These numbers are $3$ and $-1$. Rewrite the middle term: $$3x^2 + 3x - x - 1$$ Group terms: $$ (3x^2 + 3x) - (x + 1) $$ Factor each group: $$ 3x(x + 1) - 1(x + 1) $$ Factor out the common binomial: $$ (3x - 1)(x + 1) $$ 6. **Rewrite the original expression:** $$\sqrt{3(3x - 1)(x + 1)}$$ 7. **Since the square root of a product is the product of the square roots:** $$\sqrt{3} \times \sqrt{3x - 1} \times \sqrt{x + 1}$$ 8. **Final simplified form:** $$\sqrt{3} \sqrt{3x - 1} \sqrt{x + 1}$$ This is the simplified radical form, assuming $3x - 1 \geq 0$ and $x + 1 \geq 0$ for the square roots to be real.