Solve using the Quadratic Equation. 1. $$ 6 x_{2}-3 x+1=0 $$ 2. $$ x_{2}-2 x+10=0 $$ 3. $$ x_{2} \cdot 4 x+5=0 $$

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### Problem 1: $$6x^2 - 3x + 1 = 0$$

1. Identify the coefficients:  
   $$a = 6$$, $$b = -3$$, $$c = 1$$.

2. Use the quadratic formula:  
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$

3. Substitute the coefficients:
   $$
   x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(6)(1)}}{2(6)}
   $$

4. Simplify:
   $$
   x = \frac{3 \pm \sqrt{9 - 24}}{12}
   $$
   $$
   x = \frac{3 \pm \sqrt{-15}}{12}
   $$

5. Express the square root of the negative number in terms of $$i$$:
   $$
   \sqrt{-15} = i\sqrt{15}
   $$

6. Final solution:
   $$
   x = \frac{3 \pm i\sqrt{15}}{12}
   $$

---

### Problem 2: $$x^2 - 2x + 10 = 0$$

1. Identify the coefficients:  
   $$a = 1$$, $$b = -2$$, $$c = 10$$.

2. Use the quadratic formula:  
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$

3. Substitute the coefficients:
   $$
   x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)}
   $$

4. Simplify:
   $$
   x = \frac{2 \pm \sqrt{4 - 40}}{2}
   $$
   $$
   x = \frac{2 \pm \sqrt{-36}}{2}
   $$

5. Express the square root of the negative number in terms of $$i$$:
   $$
   \sqrt{-36} = 6i
   $$

6. Final solution:
   $$
   x = \frac{2 \pm 6i}{2} = 1 \pm 3i
   $$

---

### Problem 3: $$x^2 + 4x + 5 = 0$$

1. Identify the coefficients:  
   $$a = 1$$, $$b = 4$$, $$c = 5$$.

2. Use the quadratic formula:  
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$

3. Substitute the coefficients:
   $$
   x = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)}
   $$

4. Simplify:
   $$
   x = \frac{-4 \pm \sqrt{16 - 20}}{2}
   $$
   $$
   x = \frac{-4 \pm \sqrt{-4}}{2}
   $$

5. Express the square root of the negative number in terms of $$i$$:
   $$
   \sqrt{-4} = 2i
   $$

6. Final solution:
   $$
   x = \frac{-4 \pm 2i}{2} = -2 \pm i
   $$
The solutions are: 1. $$ x = \frac{3 \pm i\sqrt{15}}{12} $$ 2. $$ x = 1 \pm 3i $$ 3. $$ x = -2 \pm i $$

Solve using the Quadratic Equation.

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