Question 198603
{{{12y^2+(10)y-28}}}
Use the Quad Form
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} where a=12, b=10, c=-28
{{{12*(y+2)*(y-7/6)}}}
You got one part{{{(6y-7)}}}
The other would be {{{(2y+4)}}}

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Anthony Allie
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