[tex]\displaystyle{ fie \ \frac{2x+1}{x-3} = t, \ t \geqslant 0 }[/tex]
[tex]\displaystyle{ 2t - 3t^{-1} = 5 \rightarrow 2t - \frac{3}{t} = 5 }[/tex]
[tex]\displaystyle{ \frac{2t^{2} - 3}{t} = 5 \rightarrow 2t^{2} - 3 = 5t }[/tex]
[tex]\displaystyle{2t^{2} - 5t - 3 = 0 \rightarrow a = 2, \ b = -5, \ c = -3 }[/tex]
[tex]\displaystyle{ \Delta = b^{2} -4ac = (-5)^{2} + 4 \cdot 2 \cdot 3 }[/tex]
[tex]\displaystyle{ \Delta = 25 + 4 \cdot 6 = 25 + 24 = 49 }[/tex]
[tex]\displaystyle{ t_{1} = \frac{-b+\sqrt{\Delta}}{2a} = \frac{5 +7}{4} = \frac{12}{4} = 3 }[/tex]
[tex]\displaystyle{ t_{2} = \frac{-b-\sqrt{\Delta}}{2a} = \frac{5 - 7}{4} = \frac{-2}{4} = -0,5 }[/tex] nu convine deoarece [tex]\displaystyle{ t \geqslant 0 }[/tex]
[tex]\displaystyle{ \frac{2x+1}{x-3} = 3 \rightarrow 2x + 1 = 3x - 9 }[/tex]
[tex]\displaystyle{ x - 9 = 1 \rightarrow x = 10 }[/tex]
