Let m and n be the numbers of real roots of the quadratic equations...

Given,

${x}^{2} - 12x + \left[x\right] + 31 = 0$

$⇒ \mathop{\mathop{{x}^{2} - 12x + 31}_{⏟}}_{\ge - 5} = - \left[x\right]$

Now from above equation we say that, it could have its solution in $[ 5, 6 )$ but it does not exist as at $x = 5$ as LHS$= 1$,

So no solution, hence $m = 0$

Now solving,

${x}^{2} - 5\left|x + 2\right| - 4 = 0$

Taking Case $1$ when $x \ge - 2$ we get,

${x}^{2} - 5\left(x + 2\right) - 4 = 0$

$⇒ {x}^{2} - 5x - 14 = 0 ⇒ x = 7, - 2$

Now taking Case $2$ when $x < - 2$ we get,

${x}^{2} + 5x + 10 - 4 = 0$

$⇒ x = - 2, - 3$

So, total $3$ solution i.e., $x = - 3, - 2, 7$

Hence, $n = 3$

So, the value of ${m}^{2} + mn + {n}^{2} = 0 + 0 + {3}^{2} = 9$