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
$y =(-b+-sqrt(b^2-4ac))/2a$

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

$=2+-sqrt(21)$

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