1. **State the problem:** Find the area enclosed by the curves given by the equations $$y = 5x + 4$$ and $$y = x^2 + 2x - 6$$. 2. **Find the points of intersection:** Set the two equations equal to find the x-values where the curves intersect. $$5x + 4 = x^2 + 2x - 6$$ Rearrange to form a quadratic equation: $$x^2 + 2x - 6 - 5x - 4 = 0$$ $$x^2 - 3x - 10 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-3$$, and $$c=-10$$. $$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2}$$ $$x = \frac{3 \pm 7}{2}$$ So the two intersection points are: $$x = \frac{3 + 7}{2} = 5$$ and $$x = \frac{3 - 7}{2} = -2$$. 4. **Set up the integral for the area:** The area enclosed is the integral of the difference between the upper curve and the lower curve from $$x = -2$$ to $$x = 5$$. Since $$y = 5x + 4$$ is a line and $$y = x^2 + 2x - 6$$ is a parabola, check which is on top between the intersection points. Test at $$x=0$$: $$5(0) + 4 = 4$$ $$0^2 + 2(0) - 6 = -6$$ So, $$y = 5x + 4$$ is above $$y = x^2 + 2x - 6$$ between $$x = -2$$ and $$x = 5$$. 5. **Calculate the area:** $$\text{Area} = \int_{-2}^{5} \left[(5x + 4) - (x^2 + 2x - 6)\right] dx = \int_{-2}^{5} (-x^2 + 3x + 10) dx$$ 6. **Integrate:** $$\int (-x^2 + 3x + 10) dx = -\frac{x^3}{3} + \frac{3x^2}{2} + 10x + C$$ 7. **Evaluate the definite integral:** $$\left[-\frac{x^3}{3} + \frac{3x^2}{2} + 10x\right]_{-2}^{5} = \left(-\frac{5^3}{3} + \frac{3 \cdot 5^2}{2} + 10 \cdot 5\right) - \left(-\frac{(-2)^3}{3} + \frac{3 \cdot (-2)^2}{2} + 10 \cdot (-2)\right)$$ Calculate each part: $$-\frac{125}{3} + \frac{3 \cdot 25}{2} + 50 = -\frac{125}{3} + \frac{75}{2} + 50$$ $$= -41.6667 + 37.5 + 50 = 45.8333$$ $$-\frac{-8}{3} + \frac{3 \cdot 4}{2} - 20 = \frac{8}{3} + 6 - 20 = 2.6667 + 6 - 20 = -11.3333$$ Subtract: $$45.8333 - (-11.3333) = 45.8333 + 11.3333 = 57.1666$$ 8. **Round to 2 decimal places:** $$\boxed{57.17}$$ **Final answer:** The area enclosed by the curves is approximately 57.17 square units.