\(2x^2+6x-5\left(x+3\right)=0\)
\(\Rightarrow2x^2+6x-5x-15=0\)
\(\Rightarrow2x^2+x-15=0\)
\(\Rightarrow\left(2x^2+6x\right)-\left(5x+15\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)
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Ta có: 2x2 +6x - 5(x+3) = 0
=> 2x(x+3) -5(x+3) =0
=> (x+3)(2x -5) =0 =>\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\2x-5=0\Rightarrow x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy x\(\in\){ -3, \(\dfrac{5}{2}\)}
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