Question 1069380
You either must be given the area, or you need to assign a variable to it; then you can solve for x.


{{{A=(2x-3)(x+3)}}}

{{{2x^2-3x+6x-9-A=0}}}

{{{2x^2+3x-(A+9)=0}}}



If your value for A makes the equation factorable, then use factoring to solve.  Otherwise, use general solution for quadratic equation:


{{{x=(-3+- sqrt(3^2+4*2*(A+9)))/(2*2)}}}, from which one of the x values might not be acceptable but the other will be.