The problem is to solve the quadratic equation $55n^2 - 33n - 1940 = 0$ for the variable $n$.

1. Problem Description

The problem is to solve the quadratic equation $55n^2 - 33n - 1940 = 0$ for the variable $n$.

2. Solution Steps

We can use the quadratic formula to solve for $n$:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 55$, $b = -33$, and $c = -1940$.

Plugging these values into the quadratic formula, we get:

$n = \frac{-(-33) \pm \sqrt{(-33)^2 - 4(55)(-1940)}}{2(55)}$

$n = \frac{33 \pm \sqrt{1089 + 426800}}{110}$

$n = \frac{33 \pm \sqrt{427889}}{110}$

$n = \frac{33 \pm 654.1322}{110}$

So we have two possible values for $n$:

$n_1 = \frac{33 + 654.1322}{110} = \frac{687.1322}{110} \approx 6.2466$

$n_2 = \frac{33 - 654.1322}{110} = \frac{-621.1322}{110} \approx -5.6466$

Let's verify our results by plugging them back into the original equation.

For $n = 6.2466$:

$55(6.2466)^2 - 33(6.2466) - 1940 = 55(39.0204) - 206.1378 - 1940 = 2146.122 - 206.1378 - 1940 = 2146.122 - 2146.1378 = -0.0158$ which is close to zero.

For $n = -5.6466$:

$55(-5.6466)^2 - 33(-5.6466) - 1940 = 55(31.8843) + 186.3378 - 1940 = 1753.6365 + 186.3378 - 1940 = 1939.9743 - 1940 = -0.0257$ which is close to zero.

We can use wolfram alpha to check. The roots are $n = \frac{33 \pm \sqrt{427889}}{110}$.

$n = \frac{33 + \sqrt{427889}}{110} = \frac{33 + 654.1322}{110} \approx 6.2466$

$n = \frac{33 - \sqrt{427889}}{110} = \frac{33 - 654.1322}{110} \approx -5.6466$

3. Final Answer

The solutions are approximately $n \approx 6.2466$ and $n \approx -5.6466$.

Or in exact terms:

$n = \frac{33 \pm \sqrt{427889}}{110}$