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!

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