Question 135914
So far, so good.

{{{(1/2)x(x+3)=20}}}


{{{x(x+3)=40}}}


{{{x^2+3x-40=0}}}


Now, solve the quadratic.  You will get two solutions, one of which is negative. This is an extraneous root caused by squaring the variable in the process of setting up the equation.  Exclude the negative root.  The positive root is the value of the measure of the base.