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Δ= (-5)² -4*(-2)*3 = 25+24=49
√Δ=7
x1 = (5-7)/-4= -2/-4 = 0,5
x2 = (5+7)/ -4 = 12/ -4 = -3
x∈ (-∞; -3 )u(0,5 ; +∞)
masz nierówność postaci ax²+bx+c<0
Δ = b²-4ac = 25 -4 * (-2*3) = 25+24 = 49
√Δ=7
x1 = (-b-√Δ)/2a = (5-7)/-4 = 2/4
x2 = (-b+√Δ)/2a = (5+7)/-4 = -3
czyli x∈ (-oo:-3)V(2/4:+oo)