\(2\left(x+1\right)^2-\left(x-3\right)\left(x+2\right)=x^2-4\)
\(\Leftrightarrow2x^2+4x+2-x^2+x+6-x^2+4=0\)
\(\Leftrightarrow5x=-12\Leftrightarrow x=-\dfrac{12}{5}\)
2(x + 1)2 - (x - 3)(x + 2) = x2 - 4
<=> 2(x2 + 2x + 1) - x2 - 2x + 3x + 6 = x2 - 4
<=> 2x2 + 4x + 2 - x2 - 2x + 3x + 6 - x2 + 4 = 0
<=> 2x2 - x2 - x2 + 4x - 2x + 3x = -2 - 6 - 4
<=> 5x = -12
<=> x = \(\dfrac{-12}{5}\)