Answer
$x=15$ or $x=1$
Work Step by Step
$ 11(x-2)+(x-5)=(x+2)(x-6)\qquad$... use the distributive property: $a(b+c)=ab+ac$.
$ 11x-22+x-5=(x+2)(x-6)\qquad$... use the FOIL method.
$ 11x-22+x-5=x^{2}-6x+2x-12\qquad$...add like terms.
$ 12x-27=x^{2}-4x-12\qquad$...add $(-x^{2}+4x+12)$ to both sides.
$ 12x-27-x^{2}+4x+12=x^{2}-4x-12-x^{2}+4x+12\qquad$...add like terms.
$-x^{2}+16x-15=0\qquad$...mulitiply both sides by $-1$.
$ x^{2}-16x+15=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-16,\ c=15$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-16,\ a$ for $1$ and $c$ for $15$.
$ x=\displaystyle \frac{-(-16)\pm\sqrt{(-16)^{2}-4\cdot(15)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$x=\displaystyle \frac{16\pm\sqrt{256-60}}{2}$
$x=\displaystyle \frac{16\pm\sqrt{196}}{2}$
$ x=\displaystyle \frac{16\pm 14}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$x=\displaystyle \frac{16+14}{2}=\frac{30}{2}=15$ or $x=\displaystyle \frac{16-14}{2}=\frac{2}{2}=1$
