[tex]-x^2 -2x+1\geq 2x(x-1)-3 ~~~~~~~\text{(Aducem la o forma mai simpla)} \\
-x^2 -2x+1\geq 2x^2-2x-3 \\
-x^2 \underline{-2x}+\underline{2x}+1+3 \geq 0 \\
-3x^2 + 4 \geq 0 \\
4-3x^2 \geq 0 \\
2^2 - (x \sqrt{3})^2 \geq 0 ~~~~~~~ \text{Folosim formula: }~~ a^2 -b^2 = (a-b)(a+b)\\
(2 - x\sqrt{3})(2 + x\sqrt{3}) \geq 0 ~~~~~~~\text{Rezolvam ecuatia atasata inecuatiei}
[/tex]
[tex]2 - x\sqrt{3}=0 ~~~~~~~\Longrightarrow x_1= \frac{2}{ \sqrt{3}} = \frac{2\sqrt{3}}{3} \\
2 + x\sqrt{3}=0 ~~~~~~~\Longrightarrow x_2= \frac{-2}{ \sqrt{3}} = \frac{-2\sqrt{3}}{3}\\
Deoarece~~~inecuatia: \\
-3x^2 + 4 \geq 0 \\
\text{are coeficientul lui } ~~ x^2 \ \textless \ 0~~\text{functia atasata este pozitiva intre radacini.} \\
\Longrightarrow \boxed{ x \in \left[-\frac{2\sqrt{3}}{3},~~\frac{2\sqrt{3}}{3} \right]}[/tex]
