O Radicals and Quadratic Functions Solving a radical equation with a quadratic expression under the radical Solve for \( x \), where \( x \) is a real number. \[ x=\sqrt{x^{2}+x+6} \] If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

To solve the equation: \[ x = \sqrt{x^{2} + x + 6} \] **Step 1: Determine the Domain** The square root is defined only when the expression inside is non-negative: \[ x^{2} + x + 6 \geq 0 \] The quadratic \( x^{2} + x + 6 \) has a discriminant: \[ \Delta = 1^2 - 4 \cdot 1 \cdot 6 = 1 - 24 = -23 \] Since the discriminant is negative, \( x^{2} + x + 6 \) is always positive for all real numbers \( x \). Additionally, since the square root yields a non-negative value, \( x \) must satisfy: \[ x \geq 0 \] **Step 2: Square Both Sides** \[ x^2 = x^{2} + x + 6 \] Subtract \( x^2 \) from both sides: \[ 0 = x + 6 \implies x = -6 \] **Step 3: Check for Extraneous Solutions** The potential solution \( x = -6 \) does not satisfy the domain condition \( x \geq 0 \). Additionally, substituting \( x = -6 \) back into the original equation: \[ -6 = \sqrt{(-6)^2 + (-6) + 6} \implies -6 = \sqrt{36} \implies -6 = 6 \] This is a contradiction. **Conclusion** There are no real solutions to the equation. **Answer:** No solution