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The given equation is:

\[
x^4 - 29x^2 + 100 = 0
\]

This is a **quartic equation** but can be transformed into a **quadratic equation** by substituting \( y = x^2 \), thus making the equation:

\[
y^2 - 29y + 100 = 0
\]

Now, let's solve this quadratic equation using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1 \), \( b = -29 \), and \( c = 100 \). I'll calculate the roots now.The solutions to the quadratic equation \( y^2 - 29y + 100 = 0 \) are:

\[
y_1 = 25, \quad y_2 = 4
\]

Since we made the substitution \( y = x^2 \), we now solve for \( x \):

1. \( x^2 = 25 \) gives \( x = \pm 5 \)
2. \( x^2 = 4 \) gives \( x = \pm 2 \)

Thus, the solutions to the original equation \( x^4 - 29x^2 + 100 = 0 \) are:

\[
x = \pm 5, \quad x = \pm 2
\]

Do you want further details or have any questions?

Here are some related questions to explore:
1. How do we factor a quartic equation like this one directly?
2. What is the significance of the quadratic substitution in solving quartic equations?
3. Can quartic equations always be solved by substitution?
4. How does the discriminant help in identifying the nature of roots?
5. What are other methods to solve higher-degree polynomials?

**Tip:** For quartic equations, substitution techniques can often simplify the process, turning them into manageable quadratic forms.

Solve the equation x^4 - 29x^2 + 100 = 0.

Learn how to solve the quartic equation x^4 - 29x^2 + 100 = 0 by using quadratic substitution. This problem demonstrates key algebraic techniques and the quadratic formula. Suitable for students in Grades 10-12.

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