From Förberedande kurs i matematik 1

A first thought is perhaps to write the equation as

\displaystyle x^{2}+ax+b=0

and then try to choose the constants a and b in some way so that \displaystyle x=-1 and \displaystyle x=2 are solutions. But a better way is to start with a factorized form of a second-order equation,

\displaystyle (x+1)(x-2)=0\,\textrm{.}

If we consider this equation, we see that both \displaystyle x=-1 and \displaystyle x=2 are solutions to the equation, since \displaystyle x=-1 makes the first factor on the left-hand side zero, whilst \displaystyle x=2 makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get

\displaystyle x^{2}-x-2=0\,\textrm{.}

One answer is thus the equation \displaystyle (x+1)(x-2)=0, or \displaystyle x^{2}-x-2=0\,.

Note: There are actually many answers to this exercise, but what all second-degree equations that have \displaystyle x=-1 and \displaystyle x=2 as roots have in common is that they can be written in the form

\displaystyle ax^{2}-ax-2a=0\,,

where a is a non-zero constant.