#### Answer

The solutions are $6$ and $12$.

#### Work Step by Step

First, note that the value of $x$ cannot be $7$ as it will make some denominators equal to zero.
Multiply $(x-7)^2$ to both sides of the equation to obtain:
\begin{array}{ccc}
\require{cancel}
&(x-7)^2(1)&=&(x-7)^2\left[\dfrac{4}{x-7}+\dfrac{5}{(x-7)^2}\right]
\\&(x-7)^2&=&(x-7)^2\left(\dfrac{4}{x-7}\right)+(x-7)^2\left[\dfrac{5}{(x-7)^2}\right]
\\&(x-7)^2&=&\cancel{(x-7)^2}^{x-7}\left(\dfrac{4}{\cancel{x-7}}\right)+\cancel{(x-7)^2}\left[\dfrac{5}{\cancel{(x-7)^2}}\right]
\\&(x-7)^2&=&(x-7)(4)+5
\end{array}
Square the binomial using the formula $(a-b)^2=a^2-2ab+b^2$, distribute $4$ to the binomial, and combine like terms to obtain:
\begin{array}{ccc}
\\&x^2-2(7)(x)+7^2 &= &4(x)-7(4)+5
\\&x^2-14x+49 &= &4x-28+5
\\&x^2-14x+49 &= &4x-23
\end{array}
Move all terms to the left side of the equation.
Note that moving a term from one side to the other changes its sign.
$$x^2-14x+49-4x+23=0
\\x^2-18x+72=0$$
Factor the trinomial to obtain:
$$(x-12)(x-6)=0$$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&x-12=0 &\text{or} &x-6=0
\\&x=12 &\text{or} &x=6
\end{array}
Therefore, the solutions are $6$ and $12$.