If the set of all $a \in \mathbb{R}$, for which the equation $2x^2+(a-5)x+15=3a$ has no real root, is the interval $(\alpha, \beta)$, and $X=\{x \in \mathbb{Z}: \alpha < x < \beta\}$, then $\sum_{x \in X} x^2$ is equal to:

Step 1: Rewrite the equation in standard quadratic form

First, we need to rearrange the given equation into the standard quadratic form $Ax^2 + Bx + C = 0$. This involves moving all terms to one side of the equation. In this case, we subtract $3a$ from both sides.

Step 2: Apply the condition for no real roots

For a quadratic equation $Ax^2 + Bx + C = 0$ to have no real roots, its discriminant $D = B^2 - 4AC$ must be less than zero. Here, $A=2$, $B=(a-5)$, and $C=(15-3a)$. We substitute these values into the discriminant inequality.

Step 3: Simplify the inequality

Now, we expand and simplify the inequality obtained from the discriminant. We expand $(a-5)^2$ and distribute the terms in $4(2)(15-3a)$, then combine like terms to get a simpler quadratic inequality in terms of $a$.

Step 4: Find the roots of the quadratic inequality

To solve the quadratic inequality $a^2 + 14a - 95 < 0$, we first find the roots of the corresponding quadratic equation $a^2 + 14a - 95 = 0$. We can factor this quadratic expression to find the values of $a$ where the expression equals zero.

Step 5: Determine the interval for 'a'

Since the parabola $y = a^2 + 14a - 95$ opens upwards (coefficient of $a^2$ is positive), the expression $a^2 + 14a - 95$ is less than zero between its roots. Thus, the interval for $a$ is $(-19, 5)$. This means $\alpha = -19$ and $\beta = 5$.

Step 6: Identify integers in the interval and calculate the sum of their squares

The set $X$ consists of all integers $x$ such that $-19 < x < 5$. These integers are $\{-18, -17, \dots, 3, 4\}$. We need to calculate the sum of the squares of these integers. We can use the formula for the sum of the first $n$ squares, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. Since $x^2 = (-x)^2$, the sum from $-18$ to $4$ can be calculated by summing squares from $1$ to $18$ and then adding the squares from $1$ to $4$ and $0^2$. More simply, $\sum_{x=-18}^{4} x^2 = \sum_{x=1}^{18} x^2 - \sum_{x=1}^{4} x^2 + 0^2$. No, this is incorrect. The sum is $\sum_{x=-18}^{-1} x^2 + 0^2 + \sum_{x=1}^{4} x^2 = \sum_{x=1}^{18} x^2 + \sum_{x=1}^{4} x^2$.

Step 7: Calculate the sum of squares

Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$: \\ For $n=18$, $\sum_{x=1}^{18} x^2 = \frac{18(19)(37)}{6} = 3 \cdot 19 \cdot 37 = 2109$. \\ For $n=4$, $\sum_{x=1}^{4} x^2 = \frac{4(5)(9)}{6} = 2 \cdot 5 \cdot 3 = 30$. \\ The total sum is $2109 + 30 = 2139$.

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