Question 10269
Your set up is right.  Just solve the equation, which is a quadratic equation.  These types of equations nearly almost always factor.  You are letting x = the length of each of the two equal sides, where the 30-x represents the side parallel to the house.


x(30-2x) = 100
{{{30x - 2x^2 = 100}}}


Set the equation equal to zero, preferably with the {{{x^2}}} term positive.  To do this take everything to the right side, by adding {{{2x^2 - 30x }}} to each side of the equation.

{{{0 = 2x^2 - 30x + 100}}}  


From this, you can factor out the common factor of 2.
{{{0 = 2(x^2 - 15x + 50) }}}  


Now factor the trinomial:
{{{0 = 2(x-5)(x-10) }}}


There are two solutions:
x=5, x=10


If x=5 m, the other side is 30-2x = 20 m. 


If x = 10 m, the other side is 30-2x = 10 m.


Both solutions work in that the perimeter is 30 m, and the area is 100 m^2.


R^2 at SCC