Solve the Absolute Value Equation

Find the value of x from the absolute value equation |x² + 4x + 3| + 2x + 3 = 0
Solution to the absolute value equation
We know when solving the absolute value equations value of |x| becomes
\left | x \right |= \begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x=0 \\ x & \text{ if } x>0 \end{cases}
Before solving the problem we can rearrange the equation into
|x² + 4x + 3| = -2x – 3
Now solving the equation
We can divide the equation into two cases, that is
Case 1: x² + 4x + 3 = -2x – 3, when x² + 4x + 3 is grater than 0
Case 2: x² + 4x + 3 = 2x + 3, when x² + 4x + 3 is less than 0
From case 1
x² + 4x + 3 = -2x – 3
⇒ x² + 6x + 6 = 0
Apply quadratic formula to solve the equation
\begin{aligned} x&=\frac{-6\pm\sqrt{6^2-4\times1\times6}}{2\times1} \\ \\ &=\frac{-6\pm\sqrt{36-24}}{2} \\ \\ &=\frac{-6\pm\sqrt{12}}{2} \\ \\ \Rightarrow x&=-3\pm\sqrt{3}\\ \\ \end{aligned}
That is the solution to the problem is x = -3 + √3 and x = -3 – √3 from the case 1
Check answers. (Plug the solution and make sure they work)
When x = -3 + √3
|x² + 4x + 3| =? -2x – 3
⇒ |(-3 + √3)² + 4( -3 + √3) + 3| =? -2(-3 + √3) – 3
⇒ |9 – 6√3 + 3 – 12 + 4√3 + 3| =? 6 – 2√3 – 3
⇒ |- 2√3 + 3| =? 3 – 2√3
⇒ |- 2√3 + 3| ≠ 3 – 2√3
So, x = -3 + √3 is not a solution
When x = -3 – √3
|x² + 4x + 3| =? -2x – 3
⇒ |(-3 – √3)² + 4( -3 – √3) + 3| =? -2(-3 – √3) – 3
⇒ |9 + 6√3 + 3 – 12 – 4√3 + 3| =? 6 + 2√3 – 3
⇒ |- 2√3 + 3| =? 3 – 2√3
⇒ |- 2√3 + 3| = 3 – 2√3
So, x = -3 – √3 is a solution
From case 2
x² + 4x + 3 = 2x + 3
⇒ x² + 2x = 0
⇒ x(x + 2) = 0
so, x = 0 & x = -2
Check answers. (Plug the solution and make sure they work)
When x = 0
|x² + 4x + 3| =? -2x – 3
⇒ |0 + 0 + 3| =? 0 – 3
⇒ |3| =? -3
⇒ 3 ≠ -3
So, x = 0 is not a solution
When x = -2
|x² + 4x + 3| =? -2x – 3
⇒ |4 – 8 + 3| =? 4 – 3
⇒ |-1| =? 1
⇒ 1 = 1
So, x = -2 is a solution
That is, in general solution to the absolute value equation are x = -2 & x = -3 – √3