You have the equation:

\[
0.15x^4 - 11x^2 + 2 = 0
\]

This is a **quartic equation** in \( x \), but we can treat it as a quadratic in \( x^2 \) by making a substitution.

Let:
\[
y = x^2
\]
So the equation becomes:
\[
0.15y^2 - 11y + 2 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 0.15 \), \( b = -11 \), and \( c = 2 \).

### Step-by-Step Calculation:

1. **Discriminant**: 
\[
\Delta = b^2 - 4ac = (-11)^2 - 4(0.15)(2) = 121 - 1.2 = 119.8
\]
2. **Quadratic Formula**:
\[
y = \frac{-(-11) \pm \sqrt{119.8}}{2(0.15)} = \frac{11 \pm \sqrt{119.8}}{0.3}
\]
\[
y = \frac{11 \pm 10.95}{0.3}
\]
So, the two values of \( y \) are:
\[
y_1 = \frac{11 + 10.95}{0.3} = \frac{21.95}{0.3} = 73.17
\]
\[
y_2 = \frac{11 - 10.95}{0.3} = \frac{0.05}{0.3} \approx 0.17
\]

### Back-substitution:

Now recall that \( y = x^2 \), so:

- For \( y_1 = 73.17 \), we get \( x^2 = 73.17 \), so \( x = \pm \sqrt{73.17} \).
  \[
  x = \pm 8.55
  \]
  
- For \( y_2 = 0.17 \), we get \( x^2 = 0.17 \), so \( x = \pm \sqrt{0.17} \).
  \[
  x = \pm 0.41
  \]

### Final Solution:

The solutions for \( x \) are:
\[
x = \pm 8.55 \quad \text{or} \quad x = \pm 0.41
\]

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

---

Here are some related questions for further practice:
1. How would you solve a cubic equation?
2. What happens if the discriminant is negative in a quadratic equation?
3. Can all quartic equations be solved by substitution?
4. How do you derive the quadratic formula?
5. What is the significance of the discriminant in quadratic equations?

**Tip**: When solving higher degree polynomials, look for possible substitutions to reduce the complexity of the equation.

Discover how to solve the quartic equation 0.15x^4 - 11x^2 + 2 = 0 by using quadratic substitution. Follow the step-by-step solution, covering key concepts such as the quadratic formula and the discriminant. Ideal for students in Grades 10-12.

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