Q(u)=(7-2u+u2)-1=u2-2u+6.
x3+3x+3 has only one real root. To find it we can use Newton's iterative Method.
Let f(x)=x3+3x+3, then f'(x)=3x2+3, xn+1=xn-f(xn)/f'(xn). Let x0=0.
x1=-3/3=-1, x2=-1-(-1)/6=-⅚, x2=-449/549=-0.81785 approx, ..., x=u=-0.8177316739 approx (stable value) after a few more iterations.
Q(u)=Q(-0.8177316739)=8.304148438 approx.