Question 321366
I know there is a formula to find the roots of a quadratic equation, but I don't think there is a formula to find the roots of a cubic equation.


Just to be sure, I scoured the web and came up with an article relating to a formula for solving a cubic equation.


That article is here:


<a href = "http://www.sosmath.com/algebra/factor/fac11/fac11.html" target = "_blank">http://www.sosmath.com/algebra/factor/fac11/fac11.html</a>


While there is a formula, it appears to be a monster to apply.


Unless you're in an advanced algebra class on a college level or beyond, I doubt your problem was intended to be this complicated.


Most of the time when you are given a cubic equation, it's possible to reduce it by dividing by an integer term such as (x-1) or (x-3) or (x+1) or (x+3) but in this case neither of those worked.


I believe it's because your answer is not an integer.


Based on looking at a graph of this equation, it appears that your equation does have 3 roots.


My analysis shows the roots are:


Between x = -2.4142 and -2.4143
Between x = .4142 and .4143
Between x = .6000 and .6001


The following graph of your equation should confirm that, even though it's hard to see from the graph that there are actually two crossing points between x = .4 and x = .7


{{{graph(600,600,-3,1,-20,20,5x^3 + 7x^2 - 11x + 3)}}}


The process I used to find these roots was a mechanized process using an Excel Spreadsheet and the graph of the equation.


The graph gave me the general vicinity to look into and the spreadsheet was able to mechanize the process of finding the value at each increment of .0001 until I found where the value of y changed from plus to minus or from minus to plus.


I seriously doubt that you would have been able to find this manually without an extremely large effort.


I could be wrong.


If there's a simpler solution, please send it to me as I would be interested in  how this problem was solved other than the way that I did it.