The given equation is a quadratic equation in terms of \( x^2 \). Let's solve it step by step:

### 1. Substitute \( y = x^2 \) so the equation becomes:
\[
10y^2 - 17y + 3 = 0
\]
Now, this is a standard quadratic equation in \( y \).

### 2. Solve the quadratic equation using the quadratic formula:
The quadratic formula is:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \( 10y^2 - 17y + 3 = 0 \), the coefficients are:
- \( a = 10 \)
- \( b = -17 \)
- \( c = 3 \)

### 3. Calculate the discriminant:
\[
\Delta = b^2 - 4ac = (-17)^2 - 4(10)(3) = 289 - 120 = 169
\]
Since the discriminant is positive, there will be two real solutions.

### 4. Find the roots of the quadratic equation:
\[
y = \frac{-(-17) \pm \sqrt{169}}{2(10)} = \frac{17 \pm 13}{20}
\]

Thus, the two values for \( y \) are:
\[
y_1 = \frac{17 + 13}{20} = \frac{30}{20} = \frac{3}{2}
\]
\[
y_2 = \frac{17 - 13}{20} = \frac{4}{20} = \frac{1}{5}
\]

### 5. Substitute back \( y = x^2 \) and solve for \( x \):
For \( y_1 = \frac{3}{2} \):
\[
x^2 = \frac{3}{2} \quad \Rightarrow \quad x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2}
\]

For \( y_2 = \frac{1}{5} \):
\[
x^2 = \frac{1}{5} \quad \Rightarrow \quad x = \pm \sqrt{\frac{1}{5}} = \pm \frac{1}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5}
\]

### Final solution:
The four solutions for \( x \) are:
\[
x = \pm \frac{\sqrt{6}}{2}, \quad x = \pm \frac{\sqrt{5}}{5}
\]

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

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Here are 5 related questions to explore further:
1. How can the discriminant help determine the number of real solutions?
2. What happens when the discriminant is zero or negative?
3. Can the quadratic formula be used for any type of equation?
4. How do you simplify square roots in the final solution?
5. What are some methods for solving higher-degree polynomials?

**Tip:** Always check the discriminant first in a quadratic equation to quickly know the nature of the roots.

Learn how to solve the quartic equation 10x^4 - 17x^2 + 3 = 0 by substituting x^2 with y and applying the quadratic formula. This detailed step-by-step solution covers important algebraic concepts, suitable for students in Grades 9-12.

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