08/01/13 02:49:12 IOv5uG+w
x+3=(1/9)x^2+47/9
x-(1/9)x^2+3-47/9=0
-(1/9)x^2+x-10/3=0
ax^2+bx+c=0のときx=(-b±√(b^2-4ac))/2aから
x=(-1±√(1^2-4*-(1/9)*(10/3)))/(2*-(1/9))
x=(-1±1.57527188)/(2*-(1/9))
x≒4.5±7.08872348
あれ？
08/01/13 02:49:12 IOv5uG+w
x+3=(1/9)x^2+47/9
x-(1/9)x^2+3-47/9=0
-(1/9)x^2+x-10/3=0
ax^2+bx+c=0のときx=(-b±√(b^2-4ac))/2aから
x=(-1±√(1^2-4*-(1/9)*(10/3)))/(2*-(1/9))
x=(-1±1.57527188)/(2*-(1/9))
x≒4.5±7.08872348
あれ？