9x^2-196/3x^2-5x-42

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Solution for 9x^2-196/3x^2-5x-42 equation: 


x in (-oo:+oo)

9*x^2-((196/3)*x^2)-(5*x)-42 = 0

9*x^2+(-196/3)*x^2-5*x-42 = 0

-169/3*x^2-5*x-42 = 0

DELTA = (-5)^2-(-169/3*(-42)*4)

DELTA = -9439

DELTA < 0

x belongs to the empty set

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