To solve the system of equations by graphing, we need to find the intersection points of the following two equations:

1. $y = x^2 + 4x + 6$ (a parabola)
2. $y = -x + 2$ (a straight line)

Steps to solve:

1. Graph the parabola $y = x^2 + 4x + 6$:

   • This is a standard quadratic equation. Its vertex can be found using the formula $x = -\frac{b}{2a}$, where $a = 1$, $b = 4$, and $c = 6$.
   • The vertex is at $x = -\frac{4}{2(1)} = -2$.
   • Substituting $x = -2$ into the equation gives $y = (-2)^2 + 4(-2) + 6 = 4 - 8 + 6 = 2$. So, the vertex is at $(-2, 2)$.
   • The parabola opens upwards since $a > 0$.

2. Graph the line $y = -x + 2$:

   • This is a linear equation in slope-intercept form $y = mx + b$, where $m = -1$ (slope) and $b = 2$ (y-intercept).
   • The line passes through $(0, 2)$ and has a slope of $-1$, so another point on the line is $(1, 1)$ (moving down 1 unit and right 1 unit).

3. Find the points of intersection:

   • Set the equations equal to each other: $x^2 + 4x + 6 = -x + 2$
   • Rearrange into standard quadratic form: $x^2 + 5x + 4 = 0$
   • Factorize: $(x + 4)(x + 1) = 0$
   • Solve for $x$: $x = -4 \quad \text{and} \quad x = -1$

4. Find the corresponding $y$-values:

   • For $x = -4$: $y = -(-4) + 2 = 4 + 2 = 6$ So, one solution is $(-4, 6)$.
   • For $x = -1$: $y = -(-1) + 2 = 1 + 2 = 3$ So, the other solution is $(-1, 3)$.

Final Answer:

The solutions are: [ \boxed{(-4