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Nonreal answer if we put it into the quadratic equation 0=ax^2bx+c x=(-b[+/-](b^2-4ac)^0.5)/2a x=(7[+/-](7^2-4*3*6)^0.5)/(2*3) x=(7[+/-](49-72)^0.5)/6 x=(7[+/-]((-23)^0.5))/6 x=(7[+/-]((-23)^0.5)/6 As you cannot have the square root of a negative number (unless to take iinto account, but lets not go there) the quadratic does not cross the x-axis, so there a no real values that can be found to solve the equation (which i assume you are trying to do).

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15y ago

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