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3. Let a, b, and c be real numbers, with c +0. Show that the equation ax2 + bx + c = 0 (a) has two (different) real solutions

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a. Qu76x+e90. —- 0. = a + x + 20. 7) te laat we are here at å 20. => [m + a) / bh lace 7) not a valida > 2= -b at bayar 2x =C) ik t<час , Там - с 20. -e; 5054 ac is complex no. So, 0 has too complex conjugate solution

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