1. **State the problem:** Solve for real number $y$ in the equation $$y - 2 = \sqrt{10 - 3y}.$$\n\n2. **Understand the equation:** The right side is a square root, so the expression inside must be non-negative: $$10 - 3y \geq 0 \implies y \leq \frac{10}{3}.$$ Also, the left side $y-2$ must be non-negative because a square root is always non-negative, so $$y - 2 \geq 0 \implies y \geq 2.$$\n\n3. **Isolate and square both sides:** To eliminate the square root, square both sides: $$ (y - 2)^2 = (\sqrt{10 - 3y})^2 \implies (y - 2)^2 = 10 - 3y.$$\n\n4. **Expand and simplify:** $$ (y - 2)^2 = y^2 - 4y + 4,$$ so the equation becomes $$ y^2 - 4y + 4 = 10 - 3y.$$\n\n5. **Bring all terms to one side:** $$ y^2 - 4y + 4 - 10 + 3y = 0 \implies y^2 - y - 6 = 0.$$\n\n6. **Solve the quadratic equation:** Factor or use the quadratic formula. Factoring: $$ y^2 - y - 6 = (y - 3)(y + 2) = 0.$$\n\n7. **Find roots:** $$ y = 3 \quad \text{or} \quad y = -2.$$\n\n8. **Check solutions against domain restrictions:** Recall $2 \leq y \leq \frac{10}{3} \approx 3.33$.\n- For $y=3$: Check original equation: $$3 - 2 = 1,$$ and $$\sqrt{10 - 3(3)} = \sqrt{10 - 9} = \sqrt{1} = 1,$$ so $y=3$ is valid.\n- For $y=-2$: Check original equation: $$-2 - 2 = -4,$$ but $$\sqrt{10 - 3(-2)} = \sqrt{10 + 6} = \sqrt{16} = 4,$$ left side is $-4$, right side is $4$, not equal, so $y=-2$ is extraneous and rejected.\n\n**Final answer:** $$y = 3.$$