[tex]Area \ of \ a \ rectangle : \ \ A=240 \ m^2 \\ Perimeter \ of \ a \ rectangle : \ P=64 \ m\\\\width \ of \ a \ rectangle : \ w=? \\length\ of \ a \ rectangle: \ l=?\\\\\begin{cases}P=2w+2l \\ A=l\cdot w \end{cases}\\\\\begin{cases}64=2w+2l \\ 240=l\cdot w \end{cases}[/tex]
[tex]\begin{cases}32= w+ l \\ 24=l\cdot w \end{cases} \\\\\begin{cases} w=32- l \\ 240=l\cdot w \end{cases}\\\\substitution\\\\240=l(32-l)\\240=32l-l^2\\\l^2-32l+240=0[/tex]
[tex]a=1, \ \ b=-32, \ \ c=240 \\l_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{32-\sqrt{ (-32)^2-4 \cdot 1 \cdot 240}}{2 }=\frac{32-\sqrt{1024 -960}}{2 }=\\\\=\frac{32-\sqrt{64}}{2 }=\frac{32-8}{2}=\frac{24}{2}=12\\\\\l_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{32+\sqrt{ (-32)^2-4 \cdot 1 \cdot 240}}{2 }=\frac{32+8}{2}=\frac{40}{2}=20[/tex]
[tex]\begin{cases} l_{1}=12 \\ w=32-l \end{cases} \ \ or \ \ \begin{cases} l_{1}=20 \\ w=32-l \end{cases} \\\\ \begin{cases} l_{1}=12 \\ w=32-12 \end{cases} \ \ or \ \ \begin{cases} l_{1}=20 \\ w=32-20 \end{cases}\\ \\ \begin{cases} l_{1}=12\ m \\ w=20\ m \end{cases} \ \ or \ \ \begin{cases} l_{1}=20\ m \\ w=12 \ m \end{cases} \\\\Answer : \ The \ width \ of \ the \ lot \ is \ 12 \ m \ or \ 20 \ m .[/tex]
