1. **State the problem:** We need to solve the equation $$p(x) = (x-2)(3x-4)(2x-5) = 5$$ for $x$. 2. **Rewrite the equation:** Set the product equal to 5: $$ (x-2)(3x-4)(2x-5) = 5 $$ 3. **Expand the product:** First, multiply two factors: $$ (x-2)(3x-4) = 3x^2 -4x -6x +8 = 3x^2 -10x +8 $$ 4. **Multiply the result by the third factor:** $$ (3x^2 -10x +8)(2x-5) = 3x^2 \cdot 2x - 3x^2 \cdot 5 -10x \cdot 2x + 10x \cdot 5 + 8 \cdot 2x - 8 \cdot 5 $$ $$ = 6x^3 - 15x^2 - 20x^2 + 50x + 16x - 40 $$ 5. **Combine like terms:** $$ 6x^3 - 35x^2 + 66x - 40 $$ 6. **Set equal to 5 and bring all terms to one side:** $$ 6x^3 - 35x^2 + 66x - 40 = 5 $$ $$ 6x^3 - 35x^2 + 66x - 45 = 0 $$ 7. **Solve the cubic equation:** $$ 6x^3 - 35x^2 + 66x - 45 = 0 $$ This cubic can be solved by rational root theorem or numerical methods. 8. **Check possible rational roots:** Factors of 45 are $\pm1, \pm3, \pm5, \pm9, \pm15, \pm45$ and factors of 6 are $\pm1, \pm2, \pm3, \pm6$. Possible rational roots are $\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{3}{3}, \pm \frac{3}{6}, \pm \frac{5}{1}, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}, \pm \frac{9}{1}, \pm \frac{9}{2}, \pm \frac{9}{3}, \pm \frac{9}{6}, \pm \frac{15}{1}, \pm \frac{15}{2}, \pm \frac{15}{3}, \pm \frac{15}{6}, \pm \frac{45}{1}, \pm \frac{45}{2}, \pm \frac{45}{3}, \pm \frac{45}{6}$. 9. **Test $x=3$:** $$6(3)^3 - 35(3)^2 + 66(3) - 45 = 6(27) - 35(9) + 198 - 45 = 162 - 315 + 198 - 45 = 0$$ So, $x=3$ is a root. 10. **Divide the cubic by $(x-3)$:** Using polynomial division or synthetic division: $$6x^3 - 35x^2 + 66x - 45 \div (x-3) = 6x^2 - 17x + 15$$ 11. **Solve the quadratic:** $$6x^2 - 17x + 15 = 0$$ Use quadratic formula: $$x = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 6 \cdot 15}}{2 \cdot 6} = \frac{17 \pm \sqrt{289 - 360}}{12} = \frac{17 \pm \sqrt{-71}}{12}$$ 12. **Interpret the result:** Since the discriminant is negative, the quadratic has no real roots. 13. **Final answer:** The only real solution is $$\boxed{x = 3}$$