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### 11. \( 3x + 5 = 11 \)
We can solve this linear equation for \( x \) by isolating \( x \).

\[
3x + 5 = 11
\]
First, subtract 5 from both sides:
\[
3x = 11 - 5
\]
\[
3x = 6
\]
Now, divide by 3:
\[
x = \frac{6}{3}
\]
\[
x = 2
\]

### 12. \( x^2 - 4x + 3 = 0 \)
This is a quadratic equation. We can solve it using factoring.

First, we look for two numbers that multiply to 3 and add up to -4. These numbers are -3 and -1. So we can factor the equation as:
\[
(x - 3)(x - 1) = 0
\]
Setting each factor to 0:
\[
x - 3 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = 3 \quad \text{or} \quad x = 1
\]

So, the roots are \( x = 3 \) and \( x = 1 \).

### 13. \( 2x^2 + 3x - 2 = 0 \)
This is another quadratic equation, but it cannot be factored easily, so we will use the **quadratic formula**:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = 3 \), and \( c = -2 \).

Now, calculate the discriminant:
\[
\Delta = b^2 - 4ac = 3^2 - 4(2)(-2) = 9 + 16 = 25
\]
So, the discriminant is 25. Now we can calculate the roots:
\[
x = \frac{-3 \pm \sqrt{25}}{2(2)}
\]
\[
x = \frac{-3 \pm 5}{4}
\]
This gives two solutions:
\[
x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}
\]
\[
x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2
\]

So, the roots are \( x = \frac{1}{2} \) and \( x = -2 \).

---

### Summary:
- For equation 11: \( x = 2 \)
- For equation 12: \( x = 3 \) and \( x = 1 \)
- For equation 13: \( x = \frac{1}{2} \) and \( x = -2 \)

Do you have any questions or need more details on any part? Here are some related questions to consider:

1. How do you apply the quadratic formula in more complex problems?
2. What happens when the discriminant in the quadratic formula is negative?
3. Can every quadratic equation be factored like in equation 12?
4. How can you graphically interpret the solutions of these equations?
5. What are the different types of roots a quadratic equation can have?

**Tip**: The discriminant \( \Delta = b^2 - 4ac \) in the quadratic formula gives important information about the nature of the roots: if it's positive, there are two real roots; if it's zero, there's one real root; and if it's negative, the roots are complex.

Cari akar dari persamaan berikut:
11. 3x + 5 = 11
12. x^2 - 4x + 3 = 0
13. 2x^2 + 3x - 2 = 0

Learn how to solve a linear equation 3x + 5 = 11 and quadratic equations x^2 - 4x + 3 = 0 and 2x^2 + 3x - 2 = 0. This problem covers key algebra concepts, factoring, and using the quadratic formula. Suitable for Grades 8-10.

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