For the period 1985-2001, the average salary "y"(in thousands of dollars) per season of a Major League Baseball player can be modeled by y=7x^2-4x+392 where x is the number of years since 1985.

Write and solve an equation to find the year when the average salary was $1,904,000.

Plug in 1,904,000 for y.

1,904,000=7x^2-4x+392

Solve for X.

X= Year

I got that I need the work of solving for x. I'm also supposed to graph it to show my work.

To find the year when the average salary was $1,904,000, we need to solve the equation y = 1,904.

First, we substitute y with 1,904 in the given equation:
1,904 = 7x^2 - 4x + 392

Next, we rearrange the equation to form a quadratic equation:
7x^2 - 4x + 392 - 1,904 = 0

Simplifying and combining like terms:
7x^2 - 4x - 1,512 = 0

Now, we can solve this quadratic equation. There are several methods to solve quadratic equations, such as factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 7, b = -4, and c = -1,512. Plugging these values into the quadratic formula, we get:

x = (-(-4) ± √((-4)^2 - 4*7*(-1,512))) / (2*7)
x = (4 ± √(16 + 42,336)) / 14
x = (4 ± √42,352) / 14

Now, we can calculate the two possible solutions for x:

x = (4 + √42,352) / 14 ≈ 16.857
x = (4 - √42,352) / 14 ≈ -48.857

Since we are looking for the number of years since 1985, we discard the negative solution. Therefore, x ≈ 16.857.

To find the year, we add x to 1985:
Year = 1985 + 16.857 ≈ 2001.857

Thus, the year when the average salary was $1,904,000 is approximately 2001.