1. **Problem Statement:** Find the value of $x$ where the slope of the tangent to the curve $y = x^2 + 3x + 2$ is equal to 7. 2. **Formula and Rules:** The slope of the tangent to a curve at any point is given by the derivative of the function at that point. For a function $y = f(x)$, the slope of the tangent is $\frac{dy}{dx}$. 3. **Find the derivative:** Given $y = x^2 + 3x + 2$, differentiate with respect to $x$: $$\frac{dy}{dx} = 2x + 3$$ 4. **Set the slope equal to 7:** We want to find $x$ such that the slope of the tangent is 7: $$2x + 3 = 7$$ 5. **Solve for $x$:** $$2x = 7 - 3$$ $$2x = 4$$ $$x = \frac{4}{2} = 2$$ 6. **Answer:** The value of $x$ where the slope of the tangent is 7 is $x = 2$. This means at $x=2$, the tangent line to the curve has a slope of 7.