1. **State the problem:** Solve the equation $$x^2 + \frac{2^{x-5}}{3} - \frac{1}{3}x - \frac{2(x-25)}{15} = 2x + \frac{3}{5} + 2x + (x-1)^2$$. 2. **Rewrite and simplify the equation:** Combine like terms on the right side: $$2x + \frac{3}{5} + 2x + (x-1)^2 = 4x + \frac{3}{5} + (x-1)^2$$. 3. **Expand $(x-1)^2$:** $$ (x-1)^2 = x^2 - 2x + 1 $$ 4. **Substitute back:** $$x^2 + \frac{2^{x-5}}{3} - \frac{1}{3}x - \frac{2(x-25)}{15} = 4x + \frac{3}{5} + x^2 - 2x + 1$$ 5. **Simplify the right side:** $$4x - 2x = 2x$$ $$x^2 + \frac{2^{x-5}}{3} - \frac{1}{3}x - \frac{2(x-25)}{15} = x^2 + 2x + \frac{3}{5} + 1$$ 6. **Subtract $x^2$ from both sides:** $$\cancel{x^2} + \frac{2^{x-5}}{3} - \frac{1}{3}x - \frac{2(x-25)}{15} = \cancel{x^2} + 2x + \frac{3}{5} + 1$$ 7. **Simplify the left side term $\frac{2(x-25)}{15}$:** $$\frac{2(x-25)}{15} = \frac{2x - 50}{15}$$ 8. **Rewrite the equation:** $$\frac{2^{x-5}}{3} - \frac{1}{3}x - \frac{2x - 50}{15} = 2x + \frac{3}{5} + 1$$ 9. **Find common denominators on the left side:** The denominators are 3 and 15. Convert $\frac{1}{3}x$ to $\frac{5}{15}x$ and $\frac{2^{x-5}}{3}$ to $\frac{5 \cdot 2^{x-5}}{15}$: $$\frac{5 \cdot 2^{x-5}}{15} - \frac{5}{15}x - \frac{2x - 50}{15} = 2x + \frac{3}{5} + 1$$ 10. **Combine left side terms over 15:** $$\frac{5 \cdot 2^{x-5} - 5x - (2x - 50)}{15} = 2x + \frac{3}{5} + 1$$ 11. **Simplify numerator:** $$5 \cdot 2^{x-5} - 5x - 2x + 50 = 5 \cdot 2^{x-5} - 7x + 50$$ 12. **Rewrite equation:** $$\frac{5 \cdot 2^{x-5} - 7x + 50}{15} = 2x + \frac{3}{5} + 1$$ 13. **Multiply both sides by 15:** $$5 \cdot 2^{x-5} - 7x + 50 = 15 \cdot 2x + 15 \cdot \frac{3}{5} + 15 \cdot 1$$ 14. **Calculate right side:** $$15 \cdot 2x = 30x$$ $$15 \cdot \frac{3}{5} = 9$$ $$15 \cdot 1 = 15$$ So right side is: $$30x + 9 + 15 = 30x + 24$$ 15. **Rewrite equation:** $$5 \cdot 2^{x-5} - 7x + 50 = 30x + 24$$ 16. **Bring all terms to one side:** $$5 \cdot 2^{x-5} - 7x + 50 - 30x - 24 = 0$$ Simplify: $$5 \cdot 2^{x-5} - 37x + 26 = 0$$ 17. **Rewrite $2^{x-5}$ as $\frac{2^x}{2^5} = \frac{2^x}{32}$:** $$5 \cdot \frac{2^x}{32} - 37x + 26 = 0$$ 18. **Multiply entire equation by 32 to clear denominator:** $$5 \cdot 2^x - 37 \cdot 32 x + 26 \cdot 32 = 0$$ Calculate constants: $$37 \cdot 32 = 1184$$ $$26 \cdot 32 = 832$$ So: $$5 \cdot 2^x - 1184 x + 832 = 0$$ 19. **Final equation to solve:** $$5 \cdot 2^x = 1184 x - 832$$ 20. **Interpretation:** This transcendental equation cannot be solved algebraically in closed form. Use numerical methods (e.g., graphing, Newton-Raphson) to approximate solutions. **Summary:** The original equation simplifies to $$5 \cdot 2^x - 1184 x + 832 = 0$$ which can be solved numerically for $x$. **Answer:** The solution(s) satisfy $$5 \cdot 2^x = 1184 x - 832$$.