1. The problem is to simplify and solve the equation $$\sqrt{x} \cdot w^2 - w + 3 = 2w$$ for $w$ in terms of $x$. 2. Start by rewriting the equation: $$\sqrt{x} w^2 - w + 3 = 2w$$ 3. Move all terms to one side to set the equation to zero: $$\sqrt{x} w^2 - w + 3 - 2w = 0$$ 4. Combine like terms: $$\sqrt{x} w^2 - 3w + 3 = 0$$ 5. This is a quadratic equation in $w$: $$a = \sqrt{x}, \quad b = -3, \quad c = 3$$ 6. Use the quadratic formula: $$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 7. Substitute values: $$w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot \sqrt{x} \cdot 3}}{2 \cdot \sqrt{x}} = \frac{3 \pm \sqrt{9 - 12\sqrt{x}}}{2\sqrt{x}}$$ 8. The solutions for $w$ are: $$w = \frac{3 \pm \sqrt{9 - 12\sqrt{x}}}{2\sqrt{x}}$$ 9. Note the discriminant $9 - 12\sqrt{x}$ must be non-negative for real solutions: $$9 - 12\sqrt{x} \geq 0 \implies \sqrt{x} \leq \frac{3}{4} \implies x \leq \left(\frac{3}{4}\right)^2 = \frac{9}{16}$$ Final answer: $$w = \frac{3 \pm \sqrt{9 - 12\sqrt{x}}}{2\sqrt{x}}$$ with the domain restriction $x \leq \frac{9}{16}$ for real $w$.