A cubic determined by two points and two stationary inputs

Let $y=Ax^3+Bx^2+Cx+D$. The points give $A+B+C+D=6$ and $-8A+4B-2C+D=15$. Its derivative is $3Ax^2+2Bx+C$. Stationarity at $1/3$ gives $\tfrac13A+\tfrac23B+C=0$, equivalently $A+2B+3C=0$; at $-2$ it gives $12A-4B+C=0$.

The source GC solution is $A=2,B=5,C=-4,D=3$, so $y=2x^3+5x^2-4x+3$. Solving $y=0$ gives $x=-3.26$ to 3 significant figures, the printed answer. As an authored uniqueness check, $f^\prime(x)=2(3x-1)(x+2)$, with $f(-2)=15$ and $f(1/3)=62/27>0$, so both stationary values are positive and only the leftmost increasing section crosses the axis. The intercept coordinate is approximately $(-3.26,0)$.