Solve: (i) `27x^(2)-10x+1=0`.
(ii) `21x^(2)-28x+10=0`.

Answer

To solve the given quadratic equations, we will use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Part (i): Solve \(27x^2 - 10x + 1 = 0\)

1. Identify coefficients:
- \(a = 27\)
- \(b = -10\)
- \(c = 1\)

2. Calculate the discriminant (\(D\)):
\[ D = b^2 - 4ac = (-10)^2 - 4 \cdot 27 \cdot 1 = 100 - 108 = -8 \]

3. Since the discriminant is negative, we will have complex roots. Now, apply the quadratic formula:
\[ x = \frac{-(-10) \pm \sqrt{-8}}{2 \cdot 27} = \frac{10 \pm \sqrt{-8}}{54} \]

4. Simplify \(\sqrt{-8}\):
\[ \sqrt{-8} = \sqrt{8} \cdot i = 2\sqrt{2} \cdot i \]

5. Substitute back into the formula:
\[ x = \frac{10 \pm 2\sqrt{2}i}{54} = \frac{10}{54} \pm \frac{2\sqrt{2}i}{54} \]

6. Simplify the fractions:
\[ x = \frac{5}{27} \pm \frac{\sqrt{2}}{27}i \]

### Solution for Part (i):
\[ x = \frac{5}{27} + \frac{\sqrt{2}}{27}i \quad \text{and} \quad x = \frac{5}{27} - \frac{\sqrt{2}}{27}i \]

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### Part (ii): Solve \(21x^2 - 28x + 10 = 0\)

1. Identify coefficients:
- \(a = 21\)
- \(b = -28\)
- \(c = 10\)

2. Calculate the discriminant (\(D\)):
\[ D = b^2 - 4ac = (-28)^2 - 4 \cdot 21 \cdot 10 = 784 - 840 = -56 \]

3. Since the discriminant is negative, we will have complex roots. Now, apply the quadratic formula:
\[ x = \frac{-(-28) \pm \sqrt{-56}}{2 \cdot 21} = \frac{28 \pm \sqrt{-56}}{42} \]

4. Simplify \(\sqrt{-56}\):
\[ \sqrt{-56} = \sqrt{56} \cdot i = \sqrt{4 \cdot 14} \cdot i = 2\sqrt{14} \cdot i \]

5. Substitute back into the formula:
\[ x = \frac{28 \pm 2\sqrt{14}i}{42} = \frac{28}{42} \pm \frac{2\sqrt{14}i}{42} \]

6. Simplify the fractions:
\[ x = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i \]

### Solution for Part (ii):
\[ x = \frac{2}{3} + \frac{\sqrt{14}}{21}i \quad \text{and} \quad x = \frac{2}{3} - \frac{\sqrt{14}}{21}i \]

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