5x 4 −290x 2 +2205=0.

To solve the equation \(5x^4 - 290x^2 + 2205 = 0\), we can use a substitution to simplify it. Let's set:

\[ y = x^2 \]

Then the equation becomes:

\[ 5y^2 - 290y + 2205 = 0 \]

This is now a quadratic equation in terms of \(y\). We can solve it using the quadratic formula, which is given by:

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

Here, \(a = 5\), \(b = -290\), and \(c = 2205\). We first calculate the discriminant:

\[ D = b^2 - 4ac = (-290)^2 - 4 \cdot 5 \cdot 2205 \]

Calculating \(D\):

\[ D = 84100 - (20 \cdot 2205) = 84100 - 44100 = 40000 \]

Now, we apply the quadratic formula:

\[ y = \frac{290 \pm \sqrt{40000}}{2 \cdot 5} \] \[ \sqrt{40000} = 200 \] \[ y = \frac{290 \pm 200}{10} \]

We have two cases for \(y\):

1. \(y = \frac{490}{10} = 49\)

2. \(y = \frac{90}{10} = 9\)

Now, we revert the substitution \(y = x^2\):

1. For \(y = 49\):

\[ x^2 = 49 \implies x = \pm 7 \]

2. For \(y = 9\):

\[ x^2 = 9 \implies x = \pm 3 \]

Thus, the solutions for the original equation \(5x^4 - 290x^2 + 2205 = 0\) are:

\[ x = 7, -7, 3, -3 \]