am nevoie de ajutor rapid, doar 3)
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2x²+11x+9=0 (ax²+bx+c=0 cu a=2; b=11; c=9)
delta=b²-4ac=11²-4×2×9=49=7²
x1=(-b+radical(delta))/(2a)=(-11+7)/(2×2)=-4/4=-1
x2=(-b-radical(delta))/(2a)=(-11-7)/(2×2)=-18/4=-4,5
forma canonica:
f(x)=a(x+b/2a)²+(-delta)/4a
f(x)=2(x+11/4)²-49/8=0