1. The problem is to analyze the quadratic expression $x^2 + x + 2$. 2. The general form of a quadratic expression is $ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 3. Here, $a=1$, $b=1$, and $c=2$. 4. To understand the behavior of this quadratic, we can find its vertex and determine if it has real roots. 5. The vertex $x$-coordinate is given by the formula $$x = -\frac{b}{2a} = -\frac{1}{2 \times 1} = -\frac{1}{2}.$$ 6. Substitute $x = -\frac{1}{2}$ into the expression to find the vertex $y$-coordinate: $$y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 2 = \frac{1}{4} - \frac{1}{2} + 2 = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} = \frac{7}{4}.$$ 7. The vertex is at $\left(-\frac{1}{2}, \frac{7}{4}\right)$. 8. To check for real roots, calculate the discriminant $$\Delta = b^2 - 4ac = 1^2 - 4 \times 1 \times 2 = 1 - 8 = -7.$$ 9. Since $\Delta < 0$, the quadratic has no real roots and does not cross the $x$-axis. 10. The parabola opens upwards (because $a=1 > 0$) and its minimum value is at the vertex. Final answer: The quadratic $x^2 + x + 2$ has a vertex at $\left(-\frac{1}{2}, \frac{7}{4}\right)$ and no real roots.