1. **State the problem:** Solve the quadratic equation $$3x^2 + 5(2x + \frac{5}{3}) = 0$$. 2. **Expand the equation:** Use the distributive property to expand the term $$5(2x + \frac{5}{3})$$. $$3x^2 + 5 \times 2x + 5 \times \frac{5}{3} = 0$$ $$3x^2 + 10x + \frac{25}{3} = 0$$ 3. **Clear the fraction:** Multiply the entire equation by 3 to eliminate the denominator. $$3 \times 3x^2 + 3 \times 10x + 3 \times \frac{25}{3} = 3 \times 0$$ $$9x^2 + 30x + 25 = 0$$ 4. **Use the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a=9$$, $$b=30$$, and $$c=25$$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 30^2 - 4 \times 9 \times 25 = 900 - 900 = 0$$ 6. **Find the roots:** Since $$\Delta = 0$$, there is one real root. $$x = \frac{-30 \pm \sqrt{0}}{2 \times 9} = \frac{-30}{18}$$ Simplify the fraction: $$x = \frac{\cancel{-30}}{\cancel{18}} = -\frac{5}{3}$$ 7. **Final answer:** The equation has one real root: $$x = -\frac{5}{3}$$