# Question #7b374

Jan 8, 2018

$x \approx 0.86887 \ldots$

#### Explanation:

There is not really an algebraic way to do this, besides Newton's method.

I used Symbolab (an online math engine like Wolfram-Alpha), and got $x \approx 0.86887 \ldots$.

An alternative way would be to graph $- 6 {x}^{3} - 18 {x}^{2} - 4 x + 21 = 0$. graph{-6x^3-18x^2-4x+21 [-5.59, 4.41, -3.98, 1.02]}

If click on around $x = 0.869$, it comes up with $y = 0$, so $x = 0.869$ would also be a value.