How do you solve using the quadratic formula #(3 - y)(y + 4) = 3y - 5#?

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Answer 1 · 2015-04-29T13:09:10

First convert the given equation into a standard quadratic form by simplification
#(3-y)(y+4)=3y-5#

#3y+12-y^2-4y = 3y-5#

#-y^2 -y+12 = 3y-5#

#=-y^2 -4y+17=0#

#y^2+4y-17=0#

Now apply the quadratic formula for roots
#y =(-b+-sqrt(b^2-4ac))/2a#

#=(-4+-sqrt(16+68))/2#

#=2+-sqrt(21)#