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2x²-4x-7>-4x-12
2x²-4x-7+4x+12>0
2x²+5>0
Δ=b²-4ac
Δ=0-4*2*5=-40
brak rozwiązań
b) x(4-x)=(2x+3)(x-2)+8
4x-x²=2x²-4x+3x-6+8
4x-x²-2x²+4x-3x+6-8=0
-3x²+5x-2=0
Δ=b²-4ac
Δ=5²-4*(-3)*(-2)=25-24=1
√Δ=1
x1=-5-1/-6=-6/-6=1
x2=-5+1/-6=-4/-6=2/3