(n) $$ \quad 15-x=6 x^{2} $$ p) $$ \quad x^{3}-10 x^{2}=-9 x $$

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### Problem (n): $$15 - x = 6x^2$$
1. **Rearrange the equation** to standard quadratic form:
   $$
   6x^2 + x - 15 = 0
   $$
2. **Apply the quadratic formula**:
   $$
   x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
   $$
   where $$a = 6$$, $$b = 1$$, and $$c = -15$$.
3. **Calculate the discriminant**:
   $$
   \Delta = 1^2 - 4(6)(-15) = 1 + 360 = 361
   $$
4. **Compute the solutions**:
   $$
   x = \frac{-1 \pm \sqrt{361}}{2 \times 6} = \frac{-1 \pm 19}{12}
   $$
   - For the positive square root:
     $$
     x = \frac{-1 + 19}{12} = \frac{18}{12} = \frac{3}{2}
     $$
   - For the negative square root:
     $$
     x = \frac{-1 - 19}{12} = \frac{-20}{12} = -\frac{5}{3}
     $$

**Solutions for (n):** $$\displaystyle x = \frac{3}{2}$$ or $$\displaystyle x = -\frac{5}{3}$$.

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### Problem (p): $$x^3 - 10x^2 = -9x$$
1. **Bring all terms to one side**:
   $$
   x^3 - 10x^2 + 9x = 0
   $$
2. **Factor out the common factor $$x$$**:
   $$
   x(x^2 - 10x + 9) = 0
   $$
3. **Set each factor equal to zero**:
   - First factor:
     $$
     x = 0
     $$
   - Second factor:
     $$
     x^2 - 10x + 9 = 0
     $$
4. **Solve the quadratic equation** $$x^2 - 10x + 9 = 0$$ using the quadratic formula:
   $$
   x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2}
   $$
   $$
   x = \frac{10 \pm 8}{2}
   $$
   - For the positive square root:
     $$
     x = \frac{10 + 8}{2} = \frac{18}{2} = 9
     $$
   - For the negative square root:
     $$
     x = \frac{10 - 8}{2} = \frac{2}{2} = 1
     $$

**Solutions for (p):** $$\displaystyle x = 0$$, $$\displaystyle x = 1$$, or $$\displaystyle x = 9$$.
**Problem (n):** Solutions: $$ x = \frac{3}{2} $$ or $$ x = -\frac{5}{3} $$. **Problem (p):** Solutions: $$ x = 0 $$, $$ x = 1 $$, or $$ x = 9 $$.

(n) \( \quad 15-x=6 x^{2} \) p) \( \quad x^{3}-10 x^{2}=-9 x \)

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