General form of quadratic equation is

`ax^2+by+c=0`

And solutions to this equation are

`x_(1,2)=(-b pm sqrt(b^2-4ac))/(2a)`

Your equation is

`2x^2+(-3a-b)x+(a^2+ab)=0`

So your coefficients are

`a=2`, `b=-3a-b`, `c=a^2+ab`

In next step we put that into formula for solution of quadratic equation and calculate `x_1` and `a_2`.

For more explanations see the link below.

I hope this helps.

You need to solve the following system of equations:

`x+y=a+b`

`a/x+b/y=2`

If we use substitution method, from first equation we get

`y=a+b-x` **(1)**

Now we put that into second equation.

`a/x+b/(a+b-x)=2`

`(a(a+b-x)+bx)/(x(a+b-x))=2`

`a^2+ab-ax+bx=2ax+2bx-2x^2`

`2x^2+x(-a+b-2a-2b)+a^2+ab=0`

`2x^2+(-3a-b)x+(a^2+ab)=0`

Now we have quadratic equation.

`x_(1,2)=(3a+b pm sqrt(9a^2+6ab+b^2-8a^2-8ab))/4`` `

`x_(1,2)=(3a+b pm sqrt(a^2-2ab+b^2))/4=(3a+b pm sqrt((a-b)^2))/4=(3a+b pm(a-b))/4`

`x_1=(3a+b-a+b)/4=(a+b)/2`

`x_2=(3a+b+a-b)/4=a`

Now we put that into our substitution (1).

`y_1=a+b-x_1=a+b-(a+b)/2=(a+b)/2`

`y_2=a+b-x_2=a+b-a=b`

**So your two solutions are `(x_1,y_1)=((a+b)/2,(a+b)/2)` and `(x_2,y_2)=(a,b)`.**