quadratics are of the form: $\displaystyle ax^2 + bx + c$
you had something squared plus a constant times something plus a constant, that is in the form of the quadratic above. in this case you had $\displaystyle a = 1$, $\displaystyle x = 2y + 3$, $\displaystyle b = -5$ and $\displaystyle c = 6$. so they were in that form
and it was not an equation, you need an equal sign to be an equation
how about this problem..
-3x^3-15x^2+27x+135
i first grouped: (-3x^3-15x^2+27x)+135 = -3x(x^2+5x+9)+135
so i decided to regroup the 5x+135 = 5(x+27)
-3x(x^29) = -3x(x+3)(x-3)5(x+27)
now do i -3x+5 (x+3)(x+27)?