1. **State the problem:** Find the values of $x$ such that $p(x) = 5$ where $p(x) = 6x^3 - 35x^2 - 34x - 45$. 2. **Set up the equation:** $$6x^3 - 35x^2 - 34x - 45 = 5$$ 3. **Bring all terms to one side:** $$6x^3 - 35x^2 - 34x - 45 - 5 = 0$$ $$6x^3 - 35x^2 - 34x - 50 = 0$$ 4. **Check given roots:** The proposed roots are $x=\frac{2}{3}$, $x=4$, and $x=\frac{5}{2}$. 5. **Verify $x=\frac{2}{3}$:** $$p\left(\frac{2}{3}\right) = 6\left(\frac{2}{3}\right)^3 - 35\left(\frac{2}{3}\right)^2 - 34\left(\frac{2}{3}\right) - 45$$ $$= 6\cdot \frac{8}{27} - 35 \cdot \frac{4}{9} - 34 \cdot \frac{2}{3} - 45$$ $$= \frac{48}{27} - \frac{140}{9} - \frac{68}{3} - 45$$ Convert all to denominator 27: $$= \frac{48}{27} - \frac{420}{27} - \frac{612}{27} - \frac{1215}{27} = \frac{48 - 420 - 612 - 1215}{27} = \frac{-2199}{27} \neq 5$$ So $x=\frac{2}{3}$ is not a root of the equation $p(x)=5$. 6. **Verify $x=4$:** $$p(4) = 6(64) - 35(16) - 34(4) - 45 = 384 - 560 - 136 - 45 = -357 \neq 5$$ 7. **Verify $x=\frac{5}{2}$:** $$p\left(\frac{5}{2}\right) = 6\left(\frac{125}{8}\right) - 35\left(\frac{25}{4}\right) - 34\left(\frac{5}{2}\right) - 45$$ $$= \frac{750}{8} - \frac{875}{4} - 85 - 45 = 93.75 - 218.75 - 85 - 45 = -255 \neq 5$$ 8. **Conclusion:** The values $x=\frac{2}{3}$, $4$, and $\frac{5}{2}$ do not satisfy $p(x) = 5$. The correct roots must satisfy $$6x^3 - 35x^2 - 34x - 50 = 0$$ which can be solved using the cubic formula or numerical methods. **Final answer:** The roots are the solutions to $$6x^3 - 35x^2 - 34x - 50 = 0$$ which are not $\frac{2}{3}$, $4$, or $\frac{5}{2}$.