Use the fact that the max of x and y is greater than $\frac{x+y}{2}$ to deduce that : $M(a,b) \geq \frac{3a^2+2a+3b^2+2b}{2} = \frac{3(a+\frac{1}{3})^2 + 3(b+\frac{1}{3})^2 - \frac{2}{3} }{2}$ = $\frac{3(a+\frac{1}{3})^2 + 3(b+\frac{1}{3})^2}{2} - \frac{1}{3} \geq -\frac{1}{3}$, with equality if and only if $a=b=-\frac{1}{3}$, and so M(a,b) is minimal for these values.

$max(x,y)\ge \frac{x+y}{2}$ And do completing squares....... And assume $a^2\ge 0$ As example $a^2+23\ge 23$ always

for some unknown reason i got a = $sqrt(3)$ and b = -2