Solve for `y:9y^(4)-29y^(2)+20=0`.

Answer

To solve the equation \( 9y^4 - 29y^2 + 20 = 0 \), we can use substitution and factorization. Here are the steps:

### Step 1: Substitute \( y^2 \) with \( x \)
Let \( x = y^2 \). Then the equation becomes:
\[ 9x^2 - 29x + 20 = 0 \]

### Step 2: Factor the quadratic equation
We need to factor \( 9x^2 - 29x + 20 \). We are looking for two numbers that multiply to \( 9 \times 20 = 180 \) and add up to \(-29\). The numbers that satisfy this are \(-9\) and \(-20\).

### Step 3: Rewrite the equation
We can rewrite the equation as:
\[ 9x^2 - 9x - 20x + 20 = 0 \]

### Step 4: Group the terms
Now, we group the terms:
\[ 9x(x - 1) - 20(x - 1) = 0 \]

### Step 5: Factor by grouping
Factoring out \( (x - 1) \):
\[ (9x - 20)(x - 1) = 0 \]

### Step 6: Set each factor to zero
Now, we set each factor to zero:
1. \( 9x - 20 = 0 \)
2. \( x - 1 = 0 \)

### Step 7: Solve for \( x \)
From \( 9x - 20 = 0 \):
\[ 9x = 20 \implies x = \frac{20}{9} \]
From \( x - 1 = 0 \):
\[ x = 1 \]

### Step 8: Substitute back for \( y \)
Recall that \( x = y^2 \):
1. For \( x = \frac{20}{9} \):
\[ y^2 = \frac{20}{9} \implies y = \pm \sqrt{\frac{20}{9}} = \pm \frac{2\sqrt{5}}{3} \]
2. For \( x = 1 \):
\[ y^2 = 1 \implies y = \pm 1 \]

### Step 9: Final solutions
Thus, the solutions for \( y \) are:
\[ y = \pm 1 \quad \text{and} \quad y = \pm \frac{2\sqrt{5}}{3} \]

### Summary of the solutions:
The final solutions are:
\[ y = 1, -1, \frac{2\sqrt{5}}{3}, -\frac{2\sqrt{5}}{3} \]