Question 879337
What you suggest  you want is to factorize {{{2y^2+17y-9}}}.  This is an expression, but NOT an equation.  There are a few combinations of numbers to try, and maybe one of them will work; but only one of them at most, or maybe none of them.


(a+h)(b-k) is supposed to produce the constant term of -9.  These would be possible binomial combinations to check:


(2y-1)(y+9)

(2y+1)(y-9)

(2y-9)(y+1)

(2y+1)(y-9)

(2y+3)(y-3)

(2y-3)(y+3)


They all will produce the -9, but which one if any will reproduce the +17y?
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In fact, the first one listed gives you this:
{{{highlight_green((2y-1)(y+9))}}}---<b>This factorization works</b>.
{{{(2y-1)y+(2y-1)*9}}}
{{{2y^2-y+18y-9}}}
{{{2y^2+18y-y-9}}}
{{{2y^2+(18-1)y-9}}}
{{{highlight_green(2y^2+17y-9)}}}