Am nevoie de rezolvare repede!!
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Punctul a)
[tex]x^2+3x=0 \Rightarrow x(x+3)=0 \\ \Rightarrow x_1=0, \ x_2=-3\\ \Leftrightarrow \tt S=\{0,-3\}[/tex]
Punctul b)
[tex]4x^2-9=0\\ 4x^2=9 \Rightarrow x^2=\dfrac{9}{4} \Rightarrow x=\pm \dfrac{3}{2}\\ \Leftrightarrow \tt S=\left\{-\dfrac{3}{2}, \dfrac{3}{2}\right\}[/tex]
Punctul c)
[tex]2x^2-3x-2=0\\ x^2-\dfrac{3}{2}x-1=0\\ x^2 -2\cdot x\cdot \dfrac{3}{4}+\dfrac{9}{16}-\dfrac{25}{16}=0\\ \left(x-\dfrac{3}{4}\right)^2=\dfrac{25}{16}\\ x-\dfrac{3}{4}=\dfrac{5}{4} \ \ sau \ \ x-\dfrac{3}{4}=-\dfrac{5}{4}\\ \Rightarrow x_1=2 \ \ sau \ \ x_2=-\dfrac{1}{2} \\ \Leftrightarrow \tt S=\left\{ -\dfrac{1}{2},2 \right\}[/tex]