𝑋2−9𝑋 +14= 0
 5𝑋2−6𝑋 +1= 0
 6𝑋2−24= 0
 5𝑋2+10= 0
6𝑋2−9𝑋 = 0

Let's solve each of the quadratic equations one by one.

### 1. \( X^2 - 9X + 14 = 0 \)

This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -9 \), and \( c = 14 \). We can solve this using the **quadratic formula**:
\[
X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values:
\[
X = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)}
\]
\[
X = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2}
\]

So, the two solutions are:
\[
X = \frac{9 + 5}{2} = 7 \quad \text{or} \quad X = \frac{9 - 5}{2} = 2
\]

### 2. \( 5X^2 - 6X + 1 = 0 \)

Here, \( a = 5 \), \( b = -6 \), and \( c = 1 \). Using the quadratic formula:
\[
X = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(5)(1)}}{2(5)}
\]
\[
X = \frac{6 \pm \sqrt{36 - 20}}{10} = \frac{6 \pm \sqrt{16}}{10} = \frac{6 \pm 4}{10}
\]

Thus, the two solutions are:
\[
X = \frac{6 + 4}{10} = 1 \quad \text{or} \quad X = \frac{6 - 4}{10} = 0.2
\]

### 3. \( 6X^2 - 24 = 0 \)

We can first factor this equation:
\[
6(X^2 - 4) = 0
\]
\[
X^2 - 4 = 0
\]
\[
(X - 2)(X + 2) = 0
\]

So, the solutions are:
\[
X = 2 \quad \text{or} \quad X = -2
\]

### 4. \( 5X^2 + 10 = 0 \)

First, subtract 10 from both sides:
\[
5X^2 = -10
\]
Now, divide by 5:
\[
X^2 = -2
\]

Since \( X^2 = -2 \) has no real solution, the solutions are **imaginary**:
\[
X = \pm i\sqrt{2}
\]

### 5. \( 6X^2 - 9X = 0 \)

Factor out the common term \( X \):
\[
X(6X - 9) = 0
\]

Set each factor equal to zero:
\[
X = 0 \quad \text{or} \quad 6X - 9 = 0
\]

Solve the second equation:
\[
6X = 9 \quad \Rightarrow \quad X = \frac{9}{6} = \frac{3}{2}
\]

So, the solutions are:
\[
X = 0 \quad \text{or} \quad X = \frac{3}{2}
\]

---

### Summary of Solutions:
1. \( X^2 - 9X + 14 = 0 \): \( X = 7 \), \( X = 2 \)
2. \( 5X^2 - 6X + 1 = 0 \): \( X = 1 \), \( X = 0.2 \)
3. \( 6X^2 - 24 = 0 \): \( X = 2 \), \( X = -2 \)
4. \( 5X^2 + 10 = 0 \): \( X = \pm i\sqrt{2} \) (imaginary solutions)
5. \( 6X^2 - 9X = 0 \): \( X = 0 \), \( X = \frac{3}{2} \)

Would you like more details on any step, or do you have any questions?

---

Here are 5 related questions to explore:
1. How do we use the discriminant to determine the nature of solutions in a quadratic equation?
2. Can quadratic equations always be solved by factoring?
3. What is the geometric interpretation of the solutions to a quadratic equation?
4. How does the quadratic formula relate to completing the square?
5. What are the properties of imaginary numbers and their uses?

**Tip:** The discriminant, \( b^2 - 4ac \), determines whether the solutions to a quadratic equation are real, imaginary, or equal.

Solve the quadratic equations: X^2 - 9X + 14 = 0, 5X^2 - 6X + 1 = 0, 6X^2 - 24 = 0, 5X^2 + 10 = 0, 6X^2 - 9X = 0

Learn how to solve a variety of quadratic equations, including those with real and imaginary solutions. This detailed explanation uses the quadratic formula and factoring techniques, covering equations like X^2 - 9X + 14 = 0 and 5X^2 + 10 = 0. Suitable for Grades 9-12.

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