\(\left(2x+3\right)\left(3-2x\right)+\left(2x+3\right)^2=0\)
\(\left(2x+3\right)\left(3-2x+2x+3\right)=0\)
\(\left(2x+3\right).6=0\)
\(2x+3=0\)
\(2x=0-3\)
\(2x=-3\)
\(x=-\dfrac{3}{2}\)
Vậy \(x=-\dfrac{3}{2}\)
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=>(2x+3)(3-2x+2x+3)=0
=>2x+3=0
=>x=-3/2
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