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 $sqrt 72$ .

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

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

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

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

$y=-3+-3sqrt2$

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

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