# How do you solve x^3 + 4x^2 - x=0?

##### 1 Answer
Jul 15, 2015

I found three solutions as follows:
${x}_{1} = 0$
${x}_{2} = - 2 + \sqrt{5}$
${x}_{3} = - 2 - \sqrt{5}$

#### Explanation:

You can start collecting $x$ to get:
$x \left({x}^{2} + 4 x - 1\right) = 0$
So you get:
${x}_{1} = 0$
${x}^{2} + 4 x - 1 = 0$ that can be solved using the Quadratic Formula as:
${x}_{2 , 3} = \frac{- 4 \pm \sqrt{16 + 4}}{2}$ which gives you two additional solutions:
${x}_{2} = \frac{- 4 + \sqrt{20}}{2} = - 2 + \sqrt{5}$
and:
${x}_{3} = \frac{- 4 - \sqrt{20}}{2} = - 2 - \sqrt{5}$

Where I used the fact that $\sqrt{20} = \sqrt{5 \cdot 4} = 2 \sqrt{5}$