Question 564109
Let's write that equation in a more conventional form.
{{{x^2=8x-15}}} ---> {{{x^2-8x+15=0}}}
If you are good at factoring you immediately realize that
{{{x^2-8x+15=(x-5)(x-3)}}}
which means that the equation can be written as
{{{(x-5)(x-3)=0}}}
showing that the solutions are {{{x=5}}} and {{{x=3}}} .
If you do not like factoring, you would solve the equation in a different way (maybe apply the quadratic formula or complete the square).
No matter how you solve the equation, the solutions are {{{x=5}}} and {{{x=3}}} .
But which one is (p+3) and which one is (q-4)? No way to tell.
There will be two solution sets:
If p+3=5 and q-4=3, p=2 and q=7.
If p+3=3 and q-4=5, p=0 and q=9.