# x(x-40)(3x+2)= 240 can you find the value of X and show me how to get it

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### 2 Answers

The equation to be solved is x(x-40)(3x+2)= 240

x(x-40)(3x+2)= 240

=> (x^2 - 40x)(3x + 2) = 240

=> 3x^3 - 120x^2 + 2x^2 - 80x = 240

=> 3x^3 -198x^2 - 80x = 240

=> 3x^3 -198x^2 - 80x - 240 = 0

It is not possible to solve for x using this equation except by using the formula for roots of a cubic equation which is given at the link provided below.

As the process of working out the roots is a very tedious one I have only provided the final results.

**The roots of the equation are: 66.41, -0.2098 + i*1.077 and -0.2098 - i*1.077**

f(x) = x(x-40)(3x+2)-240

f(40) = 40(40-40)(3*40+2)-240 = -240 is negative.

f(41) = 41(41-40)(3*41+2) - 240 = 4885 is positive.

So there is a real root between 40 and 41.