https://r2.erweima.ai/mathbot/image/172613984908951hmlum9.jpg

The equation in the image is as follows:

\[
R(\vartheta) = R_{\vartheta_0} \left( 1 + \alpha_{\vartheta_0} (\vartheta - \vartheta_0) + \beta_{\vartheta_0} (\vartheta - \vartheta_0)^2 \right)
\]

You are asking to solve this equation for \(\vartheta\), or "nach \(\vartheta\) auflösen."

Let's proceed step by step to isolate \(\vartheta\).

### 1. Expand the equation:
We start by expanding the terms inside the parentheses:
\[
R(\vartheta) = R_{\vartheta_0} \left( 1 + \alpha_{\vartheta_0} (\vartheta - \vartheta_0) + \beta_{\vartheta_0} (\vartheta - \vartheta_0)^2 \right)
\]

This simplifies to:

\[
\frac{R(\vartheta)}{R_{\vartheta_0}} = 1 + \alpha_{\vartheta_0} (\vartheta - \vartheta_0) + \beta_{\vartheta_0} (\vartheta - \vartheta_0)^2
\]

### 2. Rearrange terms:
Subtract 1 from both sides to eliminate the constant term on the right-hand side:
\[
\frac{R(\vartheta)}{R_{\vartheta_0}} - 1 = \alpha_{\vartheta_0} (\vartheta - \vartheta_0) + \beta_{\vartheta_0} (\vartheta - \vartheta_0)^2
\]

Now let’s define:
\[
C = \frac{R(\vartheta)}{R_{\vartheta_0}} - 1
\]
So we have:
\[
C = \alpha_{\vartheta_0} (\vartheta - \vartheta_0) + \beta_{\vartheta_0} (\vartheta - \vartheta_0)^2
\]

### 3. Solve the quadratic equation:
This equation is quadratic in \((\vartheta - \vartheta_0)\). To make it more straightforward, define \( x = \vartheta - \vartheta_0 \). Then we have:
\[
C = \alpha_{\vartheta_0} x + \beta_{\vartheta_0} x^2
\]
Rearranging, we get:
\[
\beta_{\vartheta_0} x^2 + \alpha_{\vartheta_0} x - C = 0
\]

This is a standard quadratic equation of the form \(Ax^2 + Bx + C = 0\), where:
- \( A = \beta_{\vartheta_0} \)
- \( B = \alpha_{\vartheta_0} \)
- \( C = -\left( \frac{R(\vartheta)}{R_{\vartheta_0}} - 1 \right)\)

We can solve for \( x \) using the quadratic formula:
\[
x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]

Substitute the values:
\[
x = \frac{-\alpha_{\vartheta_0} \pm \sqrt{\alpha_{\vartheta_0}^2 - 4\beta_{\vartheta_0}(-C)}}{2\beta_{\vartheta_0}}
\]
\[
x = \frac{-\alpha_{\vartheta_0} \pm \sqrt{\alpha_{\vartheta_0}^2 + 4\beta_{\vartheta_0} C}}{2\beta_{\vartheta_0}}
\]

### 4. Final solution for \(\vartheta\):
Now, recall that \(x = \vartheta - \vartheta_0\), so:
\[
\vartheta = \vartheta_0 + \frac{-\alpha_{\vartheta_0} \pm \sqrt{\alpha_{\vartheta_0}^2 + 4\beta_{\vartheta_0} C}}{2\beta_{\vartheta_0}}
\]
where \( C = \frac{R(\vartheta)}{R_{\vartheta_0}} - 1 \).

This gives you the value of \(\vartheta\), depending on whether you take the positive or negative root.

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

### Related questions:
1. What happens when \(\beta_{\vartheta_0} = 0\) (i.e., when the quadratic term disappears)?
2. How does the behavior of \( R(\vartheta) \) change as \(\vartheta\) moves further away from \(\vartheta_0\)?
3. What is the physical or practical interpretation of the parameters \(\alpha_{\vartheta_0}\) and \(\beta_{\vartheta_0}\)?
4. How does the quadratic formula apply to solving equations in physics, specifically in angular motion?
5. What would happen if the discriminant \(\alpha_{\vartheta_0}^2 + 4\beta_{\vartheta_0} C\) is negative? What does that imply about the solution for \(\vartheta\)?

### Tip:
When solving quadratic equations, always check the discriminant (\(B^2 - 4AC\)) to determine whether the solutions are real, repeated, or complex.

Learn how to solve the quadratic equation R(θ) = Rθ₀(1 + αθ₀(θ - θ₀) + βθ₀(θ - θ₀)²) for θ. This solution involves algebraic manipulation, expanding the equation, and applying the quadratic formula to find θ. Suitable for undergraduate-level mathematics and physics students.

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