How do you find the zeros, real and imaginary, of y=x^2-3x+29 using the quadratic formula?

1 Answer
Binayaka C.
Jan 1, 2016

The roots are complex : X_1 = 1.5+5.172i and X_2 = 1.5-5.172i where i is an imaginary number (i=sqrt -1)

Explanation:

Comparing the above equation with General Quadratic equation ax^2 + bx +c we get a =1 ; b=-3; c=29 Now we see here b^2-4*a*c = -107 If b^2-4*a*c < 0 then the roots are complex number.
Roots are (-b/(2*a) + sqrt (b^2-4*a*c)/(2*a)) and (-b/(2*a) - sqrt(b^2-4*a*c)/(2*a)) or 3/2+sqrt(9-116) / 2 = 1.5+5.172i and 3/2 - sqrt(9-116) / #2 = 1.5 - 5.172i [Answer]

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