Answer: $25$

Explanation:1. Set the equations equal: Since both equations are equal to $y$, we can set them equal to each other: $x^2 + 10x + a = -0.5$

2. Rearrange into a standard quadratic form $Ax^2 + Bx + C = 0$: $x^2 + 10x + a + 0.5 = 0$ Here, $A=1$, $B=10$, and $C = a + 0.5$.

3. Condition for no real solutions: A quadratic equation has no real solutions if its discriminant ($D = B^2 - 4AC$) is less than zero. $D < 0$ $10^2 - 4(1)(a + 0.5) < 0$

4. Solve the inequality for a: $100 - 4(a + 0.5) < 0$ $100 - 4a - 2 < 0$ $98 - 4a < 0$ $98 < 4a$ $\dfrac{98}{4} < a$ $24.5 < a$

5. Determine the least possible integer value for a: The problem states that 'a' is a positive integer constant. Since $a$ must be greater than $24.5$, the least possible integer value for $a$ is $25$.

6. Conclusion: The least possible value of a is $25$.