4x+12+4/3*4x=x^2+3x

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Solution for 4x+12+4/3*4x=x^2+3x equation: 


x in (-oo:+oo)

4*x+4*(4/3)*x+12 = x^2+3*x // - x^2+3*x

4*x-x^2+4*(4/3)*x-(3*x)+12 = 0

4*x-x^2+4*(4/3)*x-3*x+12 = 0

19/3*x-x^2+12 = 0

DELTA = (19/3)^2-(-1*4*12)

DELTA = 793/9

DELTA > 0

x = ((793/9)^(1/2)-19/3)/(-1*2) or x = (-(793/9)^(1/2)-19/3)/(-1*2)

x = ((793/9)^(1/2)-19/3)/(-2) or x = ((793/9)^(1/2)+19/3)/2

x in { ((793/9)^(1/2)-19/3)/(-2), ((793/9)^(1/2)+19/3)/2 }

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