How do you find the zeros, real and imaginary, of #y=x^2+3x-9# using the quadratic formula?

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Answer 1 · 2018-08-03T06:58:15

The zeros are #x = (-3 + 3sqrt5)/2# and #x = (-3 - 3sqrt5)/2#.

#y = x^2 + 3x - 9#

The quadratic formula is #x = (-b +- sqrt(b^2 - 4ac))/(2a)#, where from our equation we know that #a = 1, b = 3, c = -9#.

Plug them into the formula:
#x = (-3 +- sqrt(3^2 - 4(1)(-9)))/(2(1))#

#x = (-3 +- sqrt(9 + 36))/2#

#x = (-3 +- sqrt(45))/2#

#x = (-3 +- 3sqrt5)/2#

Hope this helps!