1. **State the problem:** We have a function defined as $f(x) = 6x - 5$ where $x \in \mathbb{R}$. We need to find the values of $x$ for which $f(x) = x^2$. 2. **Set up the equation:** Since $f(x) = 6x - 5$, the equation becomes: $$6x - 5 = x^2$$ 3. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$x^2 - 6x + 5 = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-6$, and $c=5$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-6)^2 - 4(1)(5) = 36 - 20 = 16$$ 6. **Find the roots:** $$x = \frac{-(-6) \pm \sqrt{16}}{2(1)} = \frac{6 \pm 4}{2}$$ 7. **Calculate each root:** - For the plus sign: $$x = \frac{6 + 4}{2} = \frac{10}{2} = 5$$ - For the minus sign: $$x = \frac{6 - 4}{2} = \frac{2}{2} = 1$$ 8. **Interpretation:** The values of $x$ for which $f(x) = x^2$ are $x=1$ and $x=5$. **Final answer:** $$\boxed{x = 1 \text{ or } x = 5}$$