1. **State the problem:** Find the range of $x$ such that both inequalities hold: $$2x + 3(4 - 3x) < 8x$$ and $$6x - 5 \leq (2x - 3)^2$$ 2. **Solve the first inequality:** Expand and simplify: $$2x + 12 - 9x < 8x$$ $$-7x + 12 < 8x$$ Bring terms involving $x$ to one side: $$12 < 15x$$ Divide both sides by 15: $$x > \frac{12}{15} = \frac{4}{5}$$ 3. **Solve the second inequality:** Expand the right side: $$(2x - 3)^2 = 4x^2 - 12x + 9$$ Rewrite inequality: $$6x - 5 \leq 4x^2 - 12x + 9$$ Bring all terms to one side: $$0 \leq 4x^2 - 12x + 9 - 6x + 5$$ $$0 \leq 4x^2 - 18x + 14$$ Divide entire inequality by 2 for simplicity: $$0 \leq 2x^2 - 9x + 7$$ 4. **Find roots of quadratic:** Use quadratic formula for $2x^2 - 9x + 7 = 0$: $$x = \frac{9 \pm \sqrt{81 - 56}}{4} = \frac{9 \pm \sqrt{25}}{4}$$ $$x = \frac{9 \pm 5}{4}$$ Roots: $$x_1 = \frac{9 - 5}{4} = 1$$ $$x_2 = \frac{9 + 5}{4} = 3.5$$ 5. **Determine where quadratic is nonnegative:** Since leading coefficient $2 > 0$, parabola opens upward. So, $2x^2 - 9x + 7 \geq 0$ for $x \leq 1$ or $x \geq 3.5$. 6. **Combine both inequalities:** From first: $x > \frac{4}{5} = 0.8$ From second: $x \leq 1$ or $x \geq 3.5$ Intersection: - For $x > 0.8$ and $x \leq 1$, we get $0.8 < x \leq 1$ - For $x > 0.8$ and $x \geq 3.5$, we get $x \geq 3.5$ **Final solution:** $$x \in \left(\frac{4}{5}, 1\right] \cup [3.5, \infty)$$