1)x^2+7x+(p^2-3p+2)=0; 2)x^2-3x+(2p^2-3p-5)=0; - вопрос №16657963127232022322350332223404 от gasanova683 27.12.2020 00:31

Question

Алгебра: 1)x^2+7x+(p^2-3p+2)=0; 2)x^2-3x+(2p^2-3p-5)=0; 3)3x^2-2x+(p^2-3p-4)=0 4)x^2+7x+(4p^2-9p+5)=0​

gasanova683 · Answer

2x2−7x+3p=0D=72−4∗2∗3p=049−24p=0−24p=−49p=2449...

1)x^2+7x+(p^2-3p+2)=0; 2)x^2-3x+(2p^2-3p-5)=0; 3)3x^2-2x+(p^2-3p-4)=0 4)x^2+7x+(4p^2-9p+5)=0​