Question:

**Step 1:** Square both sides of the equation. $$ \sqrt{3x^2 + x + 5} = x - 3 $$ $$ 3x^2 + x + 5 = (x - 3)^2 $$ **Step 2:** Expand and simplify the equation. $$ 3x^2 + x + 5 = x^2 - 6x + 9 $$ $$ 2x^2 + 7x - 4 = 0 $$ **Step 3:** Solve the quadratic equation for $x$. $$ (2x - 1)(x + 4) = 0 $$ This yields two potential solutions: $x = \frac{1}{2}$ or $x = -4$. **Step 4:** Check the potential solution $x = \frac{1}{2}$ in the original equation. For $x = \frac{1}{2}$: The Left Hand Side (LHS) is: $$ \sqrt{3\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 5} = \sqrt{3\left(\frac{1}{4}\right) + \frac{1}{2} + 5} = \sqrt{\frac{3}{4} + \frac{2}{4} + \frac{20}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} $$ The Right Hand Side (RHS) is: $$ \frac{1}{2} - 3 = \frac{1}{2} - \frac{6}{2} = -\frac{5}{2} $$ Since the LHS $\left(\frac{5}{2}\right)$ does not equal the RHS $\left(-\frac{5}{2}\right)$, $x = \frac{1}{2}$ is an extraneous solution. Furthermore, the RHS of the original equation, $x-3$, must be non-negative for a real square root to equal it. For $x=\frac{1}{2}$, $x-3 = -\frac{5}{2} < 0$, confirming it is extraneous. **Step 5:** Check the potential solution $x = -4$ in the original equation. For $x = -4$: The LHS is: $$ \sqrt{3(-4)^2 + (-4) + 5} = \sqrt{3(16) - 4 + 5} = \sqrt{48 - 4 + 5} = \sqrt{49} = 7 $$ The RHS is: $$ -4 - 3 = -7 $$ Since the LHS $(7)$ does not equal the RHS $(-7)$, $x = -4$ is an extraneous solution. **Conclusion:** Both potential solutions, $x = \frac{1}{2}$ and $x = -4$, are extraneous. Therefore, the original equation has no real solution.