Answer:
[tex]W=(x-6)\ m[/tex]
Step-by-step explanation:
we know that
The area of rectangle is given by the formula
[tex]A=LW[/tex]
we have
[tex]A=(x^2-4x-12)\ m^2[/tex]
[tex]L=(x+2)\ m[/tex]
substitute
[tex]x^2-4x-12=(x+2)W[/tex]
Solve the quadratic equation of the left side
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^2-4x-12=0[/tex]
so
[tex]a=1\\b=-4\\c=-12[/tex]
substitute in the formula
[tex]x=\frac{-(-4)\pm\sqrt{-4^{2}-4(1)(-12)}} {2(1)}[/tex]
[tex]x=\frac{4\pm\sqrt{64}} {2}[/tex]
[tex]x=\frac{4\pm8} {2}[/tex]
[tex]x=\frac{4+8} {2}=6[/tex]
[tex]x=\frac{4-8} {2}=-2[/tex]
therefore
[tex]x^2-4x-12=(x+2)(x-6)[/tex]
substitute in the formula of area
[tex](x+2)(x-6)=(x+2)W[/tex]
solve for W
simplify
[tex]W=(x-6)\ m[/tex]
