The length of a rectangle is 5 greater than 3 times the width. The area of the rectangle is 22cm. Find the length and the width. (SHOW STEPS)

$$x-width\\3x+5-length\\22cm^2-area\ of\ the\ rectangle\\\\x\cdot(3x+5)=22\\3x^2+5x-22=0\\\\a=3;\ b=5;\ c=-22\\\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot3\cdot(-22)=25+264=289\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{289}=17$$

$$x_1=\frac{-5-17}{2\cdot3}=\frac{-22}{6} < 0\\\\x_2=\frac{-5+17}{2\cdot3}=\frac{12}{6}=2\\\\3x+5=3\cdot2+5=6+5=11\\\\Answer: the\ width\ is\ 2cm\ and\ the\ lenght\ is\ 11cm.$$

$$a-the\ length\ of\ a\ rectangle,\ \ \ a>0\\b-the\ width\ of\ a\ rectangle,\ \ \ b>0\\\\a=3b+5\ \ \ \wedge\ \ \ A=22\ cm^2\\ \\A=a\cdot b\ \ \ \Leftrightarrow\ \ \ (3b+5)\cdot b=22\ \ \Leftrightarrow \ \ 3b^2+5b-22=0\\ \\\Delta=5^2-4\cdot3\cdot(-22)=25+264=289\ \ \Rightarrow\ \ \sqrt{\Delta} =17\\ \\b_1= \frac{-5-17}{2\cdot3} = \frac{-22}{6}=-3 \frac{2}{3} <0,\ \ \ \ \ \ \ \ \ b_2= \frac{-5+17}{2\cdot3} = \frac{12}{6}=2\\ \\b=2\ \ \ \Rightarrow\ \ \ \ a=3 b+5=3\cdot2+5=6+5=11$$

$$Ans. \ the\ length\ is\ 11\ cm,\ \ the\ width\ is\ 2\ cm.$$

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