[tex] {ax}^{2} +b x + c = 0 \\ D > 0 \\ D = {b}^{2} - 4ac \\ {b}^{2} - 4ac > 0[/tex]
Ответ: 3)
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[tex] {ax}^{2} +b x + c = 0 \\ D > 0 \\ D = {b}^{2} - 4ac \\ {b}^{2} - 4ac > 0[/tex]
Ответ: 3)