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

Jul 14, 2016

$x = \left\{- 3 , - 1 , 1\right\}$

#### Explanation:

First making $y = {x}^{2} + 2 x$ and substituting

${y}^{2} - 2 y - 3 = 0$

Solving for $y$ gives $y = \left\{- 1 , 3\right\}$

The next step is to solve

1) ${x}^{2} + 2 x = - 1$ giving two equal roots $x = \left\{- 1 , - 1\right\}$
2) ${x}^{2} + 2 x = 3$ giving $x = \left\{- 3 , 1\right\}$

The solutions are

$x = \left\{- 3 , - 1 , 1\right\}$

also we have

${\left({x}^{2} + 2 x\right)}^{2} - 2 \left({x}^{2} + 2 x\right) - 3 = \left(x + 3\right) {\left(x + 1\right)}^{2} \left(x - 1\right)$