Let x and y be the dimension of the rectangle. Suppose x is the width and y is the length. Since the perimeter is 36, we have
[tex] 2(x+y)=36 \iff x+y=18 \iff y = 18-x [/tex]
If we increase the length by 1, we change [tex] y\mapsto y+1 [/tex]. Similarly, if the width is increased by 2, we change [tex] x\mapsto x+2 [/tex]. So, the increased area is
[tex] (x+2)(y+1)=30 [/tex]
If we substitute the value for y deduced in the first equation, we have
[tex] (x+2)(18-x+1)=30 \iff (x+2)(19-x) = 30 \iff -x^2+17x+38 = 30 \iff x^2-17x-8=0[/tex]
The solutions to this equation are
[tex] \dfrac{17\pm\sqrt{321}}{2} [/tex]
One of them is negative, so we have to choose the positive one:
[tex] x=\dfrac{17+\sqrt{321}}{2} [/tex]
Which implies
[tex] y = 18 - \dfrac{17+\sqrt{321}}{2} = \dfrac{19-\sqrt{321}}{2} [/tex]
