Start by writing the system down, I will use [tex]y[/tex] to represent [tex]f(x)[/tex]
[tex]y=x^2-5x+3\wedge y=-3[/tex]
Substitute the fact that [tex]y=-3[/tex] into the first equation to get,
[tex]-3=x^2-5x+3[/tex]
Simplify into a quadratic form ([tex]ax^2+bx+c=0[/tex]),
[tex]x^2-5x+6=0[/tex]
Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,
[tex]x^2+x(m+n)+mn=0[/tex]
which then must factor into
[tex](x+m)(x+n)=0[/tex]
And the solutions will be [tex]m,n[/tex].
Clearly for small coefficients like ours [tex]5,6[/tex], this is very easy to figure out. To get 5 and 6 we simply say that [tex]m=3, n=2[/tex].
This fits the definition as [tex]5=3+2[/tex] and [tex]6=2\cdot3[/tex].
So as mentioned, solutions will equal to [tex]m=3, n=2[/tex] but these are just x-values in the solution pairs of a form [tex](x,y)[/tex].
To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.
So the solution pairs are [tex](2,0)[/tex] and [tex](3,0)[/tex].
Hope this helps :)
