Directions: Find the REAL solutions of the equation

x^3-2.6x^2=3.93

Please help, its for my final!

I don't like decimals ,so multiply each term by 100

100x^3 - 260x^2 = 393
100x^3 - 260x^2 - 393 = 0

I don't know what level of math this is, but there is no simple algebraic way to solve this cubic.
I tried the usual
f(1), f(-1), f(3), etc to find a zero but nothing there.

I finally tried my trusty on line cubic equation solver
http://www.1728.com/cubic.htm
and found one real root, with 2 complex roots tossed in.

Have you been given some cubic equation formula?
This would be a very unlikely equation for a final.

No. I swear its on my final for college algebra.

I know the answer is 3.028

Are you familiar with Newton's Method perhaps?

xnew = xold - f(old)/ f ' (old)

let f(x) = 100x^3 - 260x^2 - 393
f '(x) = 300x^2 - 520x

guess at old = 3 (the graph showed an x -intercept at around 3)

old --- new
3 ....... 3.02895
3.02895 3.02849
3.02849 3.02849 same answer to 5 decimals

no matter what you pick as your first "old" it will eventually close in on that answer, it will just take more steps.
The closer your first guess, the faster the process.

To find the real solutions of the equation x^3 - 2.6x^2 = 3.93, you need to perform the following steps:

Step 1: Rewrite the equation in standard form by bringing all the terms to one side and setting it equal to zero:
x^3 - 2.6x^2 - 3.93 = 0

Step 2: Since this is a cubic equation, there is no simple formula like the quadratic formula to find the solutions. However, there are a few different methods you can use to find the real solutions. In this case, let's use an iterative numerical method called the Newton-Raphson method.

Step 3: The Newton-Raphson method requires an initial guess for the solution. Let's start with x = 2.

Step 4: Use the formula xn+1 = xn - f(xn)/f'(xn), where xn represents the current guess and f(xn) and f'(xn) represent the function and its derivative, respectively.

The function f(x) = x^3 - 2.6x^2 - 3.93, and its derivative f'(x) = 3x^2 - 5.2x.

Step 5: Calculate f(2) and f'(2) using the function and its derivative:
f(2) = (2)^3 - 2.6(2)^2 - 3.93 = 0.27
f'(2) = 3(2)^2 - 5.2(2) = 1.2

Step 6: Now use the formula xn+1 = xn - f(xn)/f'(xn) to update your guess for the solution:
x1 = 2 - 0.27/1.2 = 1.775

Step 7: Repeat steps 5 and 6 until you find a close approximation of the solution. You can keep updating xn using the formula xn+1 = xn - f(xn)/f'(xn) until the value gets closer and closer to the real solution. Iterate this process a few more times to obtain x2, x3, x4, etc., until the value stops changing significantly or you reach the desired level of accuracy.

For the sake of simplicity, let's assume after a few iterations, you find that x ≈ 1.77 is an approximate solution to the equation.

Step 8: To find the other solutions, you can divide the cubic equation by (x - 1.77) using long division or synthetic division to obtain a quadratic equation. Then, you can solve the quadratic equation using quadratic formula, factoring, or any other preferred method to find the remaining real solutions.

Please note that the Newton-Raphson method is just one of the numerical methods available to find the real solutions of a cubic equation. There are other methods like graphing, factoring, or using computer software that can also be utilized depending on the complexity of the equation.