1. **State the problem:** Solve the quadratic equation $$2x(x + 4) = -3x^2 - 2x + 40$$ for $x$. 2. **Rewrite the equation:** Expand the left side: $$2x(x + 4) = 2x^2 + 8x$$ So the equation becomes: $$2x^2 + 8x = -3x^2 - 2x + 40$$ 3. **Bring all terms to one side:** Add $3x^2 + 2x - 40$ to both sides: $$2x^2 + 8x + 3x^2 + 2x - 40 = 0$$ Simplify: $$5x^2 + 10x - 40 = 0$$ 4. **Simplify the equation:** Divide all terms by 5: $$x^2 + 2x - 8 = 0$$ 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=2$, $c=-8$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36$$ 7. **Find the roots:** $$x = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$$ 8. **Calculate each root:** - $$x_1 = \frac{-2 + 6}{2} = \frac{4}{2} = 2$$ - $$x_2 = \frac{-2 - 6}{2} = \frac{-8}{2} = -4$$ **Final answer:** The solutions to the quadratic equation are $$x = 2$$ and $$x = -4$$.