If the set of all a$\in$R, for which the equation 2x² + (a – 5)x + 15 = 3a has no real root, is the interval ($\mathrm{\alpha },\mathrm{\beta }$), and X = {x $\in$ Z : $\mathrm{\alpha }$ < x < $\mathrm{\beta }$}, then $\sum _{\mathrm{x}\in \mathrm{X}} {\mathrm{x}}^2$ is equal to

Since 2x² + (a – 5)x + 15 = 3a has no real root then

(a – 5)² – 8(15 – 3a) < 0
a² + 14a + 25 – 120 < 0
a² + 14a – 95 < 0
(a + 19)(a – 5) < 0
a$\in$(–19, 5)
$\therefore$ –19 < x < 5
$\begin{array}{l}\therefore \sum _{\mathrm{x}\in \mathrm{X}} {\mathrm{x}}^2=\left(1^2+2^2+\dots +4^2\right)+\left(1^2+2^2+\dots +{18}^2\right)\\ =\frac{4\times 5\times 9}{6}+\frac{18\times 19\times 37}{6}\end{array}$
= 30 + 2109
= 2139