Law of quadratic insanity

$\large y=\frac{ax^2-7x+5}{5x^2-7x+a}$

$(5y-a)x^2-7(y-1)x+(ay-5)=0$

This is quadratic in x and must have real roots

i.e.Discriminant must be nonnegative.

$49(y-1)^2-4(5y-a)(ay-5)\geq 0$

$(49-20a)y^2+(2+4a^2)y+(49-20a)\geq 0$

This above inequality is true for all $y\in R$ as it is given that the expression is capable of all values.

So $49-20a>0$ and Discriminant <0

$a<\frac{49}{2}\quad (1)$

$4(1+2a^2)^2-4(49-20a)^2<0$

$(a^2-10a+25)(a^2-10a+24)<0$

$(a-5)^2(a^2-10a+24)<0$

$a^2-10a+24<0$

$a\in (-12,2)$ which also satisfies the inequality (1) obtained above

So $p=-12,q=2$