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x(2x+1)=3
2x²+x=3
2x²+x-3=0
Δ=b²-4ac
Δ=1+4*2*3
Δ=25 √Δ=5
x1=(-b-√Δ)/2a x2=(-b+√Δ)/2a
x1=(-1-5)/2*2 x2=(-1+5)/4
x1=-6/4 x2=1
x1=-3/2
musimy jeszcze zrobić założenie że 2x+1≠0
2x≠-1
x≠-1/2