Answer
$x=5$
Dimensions of the rectangle: $13\,m$ and $7\,m$.
Work Step by Step
$\text{length}\times\text{width}=A$
$\implies (2x+3)(x+2)=91$
$\implies 2x^{2}+7x+6=91$
$\implies 2x^{2}+7x+6-91=0$
$\implies 2x^{2}+7x-85=0$
Identfy $a$, $b$, $c$ from the standard form of a quadratic equation:
$ax^2+bx+c=0$
$a=2, b=7$ and $c=-85$.
Apply the Quadratic Formula to find the solutions:
$x=\frac{-b\pm \sqrt {b^{2}-4ac}}{2a}$
$=\frac{-7\pm \sqrt {7^{2}-4(2)(-85)}}{2(2)}$
$\implies x=5$ or $x=-\frac{17}{2}$
Since $\text{length}$ and $\text{width}$ cannot be negative, $x$ should be equal to $5$.
Dimensions of the rectangle are $(2x+3)m=[2(5)+3]\,m=13\,m$
$(x+2)m=(5+2)m=7\,m$
