Solving a quadratic equation with 3 parameters

Question

me and my group of students are having trouble solving the following quadratic equation. Any help is appreciated. Thanks in advance.

Answer 1

$$x^2-bx+6a^2+5ab=5ax+6b^2 \implies x^2+(-b-5a)x+6a^2+5ab-6b^2$$ $$ \implies x=\frac{(b+5a)\pm(a-5b)}{2}\implies x=3a-2b, 2a+3b $$ Here we have used the fact that roots of $Ax^2+Bx+C=0$ are $\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

Answer 2

What if $a=0?$

Otherwise, $$x^2-x(5a+b)+6a^2+5ab-6b^2=0$$

$$x=\dfrac{5a+b\pm\sqrt{(5a+b)^2-4(6a^2+5ab-6b^2)}}2$$

Now the discriminant is $$(5a+b)^2-4(6a^2+5ab-6b^2)=a^2-10ab+25b^2=(a-5b)^2$$