You need to solve system of equations

`{(xy=90),(2x+y=36):}`

From second equation we have

`y=36-2x`. **(1)**

So we put that into first equation instead of `y`.

`x(36-2x)=90`

`36x-2x^2=90`

`-2x^2+36x-90=0`

`x^2-18x-45=0`

We can wow solve this quadratic equation by using this formula` `

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

where `a, b` and `c` are coefficients of equation `ax^2+bx+c=0`.

So in your case

`x_(1,2)=(18pm sqrt(324-180) )/2=(18pm12)/2`

`x_1=3`, `x_2=15`

Now by putting this into (1) we get

`y_1=36-2x_1=36-6=30`

`y_2=36-2x_2=36-30=6`

**So your solutions are** `(x_1,y_1)=(3,30)` **and** `(x_2,y_2)=(15,6)`**.**