How do you solve #y^2 +6y = 9# using the quadratic formula?

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Answer 1 · 2015-11-08T06:00:20

#y=-3(1-sqrt 2), -3(1+sqrt 2)#

Move all terms to the left side.

#y^2+6y-9=0# is a quadratic equation, #ax^2+bx+c#, where #a=1, b=6, and c=-9#.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Substitute #y# for #x#.

#y=(-6+-sqrt(6^2-4*1*-9))/(2*1)#

Simplify.

#y=(-6+-sqrt(36+36))/(2)=#

#y=(-6+-sqrt72)/2#

Factor #sqrt 72#.

#y=(-6+-sqrt(2xx2xx2xx3xx3))/2=#

#y=(-6+-sqrt(2xx2^2xx3^2))/2=#

#y=(-6+-2xx3sqrt2)/2=#

#y=(-6+-6sqrt2)/2=#

#y=-3+-3sqrt2#

#y=-3(1-sqrt 2), -3(1+sqrt 2)#