Answer:
The solution is [tex]x=6\ cm[/tex]
Step-by-step explanation:
we know that
The area of rectangle is equal to
[tex]A=LW[/tex]
we have
[tex]A=56\ cm^{2}[/tex]
so
[tex]56=LW[/tex] ------> equation A
[tex]L=x+2[/tex] -----> equation B
[tex]W=2x-5[/tex] -----> equation C
Substitute equation B and equation C in equation A
[tex]56=(x+2)(2x-5)[/tex]
solve for x
[tex]56=(x+2)(2x-5)\\56=2x^{2}-5x+4x-10\\2x^{2}-x-66=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^{2}-x-66=0[/tex]
so
[tex]a=2\\b=-1\\c=-66[/tex]
substitute in the formula
[tex]x=\frac{1(+/-)\sqrt{-1^{2}-4(2)(-66)}} {2(2)}[/tex]
[tex]x=\frac{1(+/-)\sqrt{529}} {4}[/tex]
[tex]x1=\frac{1(+)23} {4}=6[/tex]
[tex]x2=\frac{1(-)23} {4}=-5.5[/tex]
The solution is [tex]x=6\ cm[/tex]
[tex]L=6+2=8\ cm[/tex]
[tex]W=2(6)-5=7\ cm[/tex]
