How do you solve #y=(5(10-y))/(3(y+1))# using the quadratic formula?

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Answer 1 · 2016-07-03T01:34:23

#y = (50 - 5y)/(3y + 3)#

#y(3y + 3) = 50 - 5y#

#3y^2 + 3y = 50 - 5y#

#3y^2 + 8y - 50 = 0#

#y = (-8 +- sqrt(8^2 - 4 xx 3 xx -50))/(2 xx 3)#

#y = (-8 +- sqrt664)/6#

#y= (-8 +- 4sqrt166)/6#

#y = (-4 +- 2sqrt166)/3#

Hopefully this helps!