1. Solve $x^2 - 9x + 14 = 0$ by factoring: Factor the quadratic: $x^2 - 9x + 14 = (x-7)(x-2) = 0$ Set each factor to zero: $x-7=0$ or $x-2=0$ Solution: $x=7$ or $x=2$ 2. Solve $x^2 - 18 = 6x$ by rearranging and factoring or quadratic formula: Write in standard form: $x^2 - 6x - 18 = 0$ Use quadratic formula: $x=\frac{6 \pm \sqrt{(-6)^2 - 4(1)(-18)}}{2(1)}=\frac{6 \pm \sqrt{36 + 72}}{2}=\frac{6 \pm \sqrt{108}}{2}$ Simplify: $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$ So, $x = \frac{6 \pm 6\sqrt{3}}{2} = 3 \pm 3\sqrt{3}$ 3. Solve $a x^2 + 2 d x - 3 c = 0$ using quadratic formula: $a x^2 + 2 d x - 3 c = 0$ Here, $A = a$, $B = 2d$, $C = -3c$ Apply formula: $$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-2d \pm \sqrt{(2d)^2 - 4a(-3c)}}{2a} = \frac{-2d \pm \sqrt{4d^2 + 12ac}}{2a}$$ Simplify under root: $\sqrt{4d^2 + 12ac} = 2\sqrt{d^2 + 3ac}$ Therefore, $$x = \frac{-2d \pm 2\sqrt{d^2 + 3ac}}{2a} = \frac{-d \pm \sqrt{d^2 + 3ac}}{a}$$ 4. Solve $x^2 + (3cx + ax) + 3ac = 0$: Rewrite: $x^2 + x(3c + a) + 3ac = 0$ Factor: Look for two numbers that multiply to $3ac$ and sum to $3c + a$, those are $3c$ and $a$ Thus, $(x + a)(x + 3c) = 0$ Set each factor to zero: $x + a = 0$ or $x + 3c = 0$ Solutions: $x = -a$ or $x = -3c$ 5. Solve $1 = 12x - 9x^2$: Rewrite in standard form: $-9x^2 + 12x - 1 = 0$ Multiply the entire equation by $-1$ to simplify: $9x^2 - 12x + 1 = 0$ Use quadratic formula: $x = \frac{12 \pm \sqrt{(-12)^2 - 4(9)(1)}}{2(9)} = \frac{12 \pm \sqrt{144 - 36}}{18} = \frac{12 \pm \sqrt{108}}{18}$ Simplify $\sqrt{108}$: $6\sqrt{3}$ So, $x = \frac{12 \pm 6\sqrt{3}}{18} = \frac{2 \pm \sqrt{3}}{3}$ Answer summary: 1. $x=7$ or $x=2$ 2. $x=3 \pm 3\sqrt{3}$ 3. $x=\frac{-d \pm \sqrt{d^2 + 3ac}}{a}$ 4. $x=-a$ or $x=-3c$ 5. $x=\frac{2 \pm \sqrt{3}}{3}$