[tex]\displaystyle (x+2)(x-2)+4(x+3)=(x+2)^2-3x \\ x^2-4+4x+12=x^2+2 \cdot x \cdot 2+2^2-3x \\ x^2-4+4x+12=x^2+4x+4-3x \\ x^2-x^2+4x-4x+3x=4+4-12 \\ 3x=-4 \Rightarrow x=- \frac{4}{3} [/tex]
[tex]\displaystyle Verificare : \\ \left(- \frac{4}{3} +2\right)\left(- \frac{4}{3} -2\right)+4\left(- \frac{4}{3} +3 \right)=\left(- \frac{4}{3} +2 \right)^2-3 \cdot \left(- \frac{4}{3} \right) \\ \\ \frac{16}{9} -4- \frac{16}{3} +12=\left(- \frac{4}{3}+\frac{2}{3} \right)^2+ \frac{12}{3} \\ \\ \frac{16-4 \cdot 9-16 \cdot 3+12\cdot 9 }{9}=\left(- \frac{2}{3} \right)^2+4\\ \\ \frac{16-36-48+108}{9} = \frac{4}{9} +4[/tex]
[tex]\displaystyle \frac{40}{9} = \frac{4+4 \cdot 9}{9}\\ \\ \frac{40}{9} = \frac{4+36}{9}\\ \\\frac{40}{9} = \frac{40}{9} \rightarrow Adevarat \\ \\ Deci,~ \boxed{x=- \frac{4}{3} }[/tex]
