Решите пожалуйста пример прошу

Решите задачу:

$\displaystyle |x^2-4x-6|=2x+3\\\\\left[\begin{array}{ccc}x^2-4x-6=-(2x+3)\\x^2-4x-6=2x+3\end{array}\right\\\\\\x^2-4x-6=-(2x+3)\\x^2-4x-6=-2x-3\\x^2-4x+2x-6+3=0\\\\x^2-2x-3=0\\\\D=b^2-4ac=(-2)^2-4\cdot1\cdot(-3)=4+12=16\\\\x_{1,2}=\frac{-b\underline+\sqrt{D}}{2a}\\\\x_1=\frac{2+\sqrt{16}}{2}=\frac{2+4}{2}=\frac62=3\\\\x_2=\frac{2-\sqrt{16}}{2}=\frac{2-4}{2}=\frac{-2}2=-1$

$\displaystyle x^2-4x-6=2x+3\\x^2-4x-2x-6-3=0\\\\x^2-6x-9=0\\\\D=b^2-4ac=6^2-4\cdot1\cdot(-9)=36+36=72\\\\x_{1,2}=\frac{-b\underline+\sqrt{D}}{2a}\\\\\sqrt{72}=\sqrt{36\cdot2}=6\sqrt2\\\\x_1=\frac{6+\sqrt{72}}{2}=\frac{6+6\sqrt{2}}{2}=3+3\sqrt{2}\\\\x_2=\frac{6-\sqrt{72}}{2}=\frac{6-6\sqrt{2}}{2}=3-3\sqrt{2}$

$\left[\begin{array}{ccc}x_1=3\\x_2=-1\\x_3=3+3\sqrt{2}\\x_4=3-3\sqrt{2}\end{array}\right\;\Longrightarrow\;\;\;\left[\begin{array}{ccc}x_1=3-3\sqrt2\\x_2=-1\;\;\;\;\;\;\;\;\\x_3=3\;\;\;\;\;\;\;\;\;\;\;\\x_4=3+3\sqrt2\end{array}\right$