Given,
x2−12x+[x]+31=0
⇒≥−5⏟x2−12x+31=−[x]
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,
x2−5∣x+2∣−4=0
Taking Case 1 when x≥−2 we get,
x2−5(x+2)−4=0
⇒x2−5x−14=0⇒x=7,−2
Now taking Case 2 when x<−2 we get,
x2+5x+10−4=0
⇒x=−2,−3
So, total 3 solution i.e., x=−3,−2,7
Hence, n=3
So, the value of m2+mn+n2=0+0+32=9