#### Answer

no solution

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{x^2+12x-4}=x-4
,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation.
$\bf{\text{Solution Details:}}$
Squaring both sides of the equation results to
\begin{array}{l}\require{cancel}
\left( \sqrt{x^2+12x-4} \right)^2=(x-4)^2
\\\\
x^2+12x-4=(x-4)^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^2+12x-4=(x)^2-2(x)(4)+(4)^2
\\\\
x^2+12x-4=x^2-8x+16
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
(x^2-x^2)+(12x+8x)=16+4
\\\\
20x=20
\\\\
x=\dfrac{20}{20}
\\\\
x=1
.\end{array}
Upon checking, $
x=1
$ DOES NOT satisfy the original equation. Hence, there is $\text{
no solution
.}$