Answer
$x=\left\{ -1,3 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{6x+7}-9=x-7
,$ use the properties of equality to isolate the radical and then square both sides. Then use the properties of equality to express the resulting equation in the form $ax^2+bx+c=0.$ Next is to factor the quadratic equation and solve for the value/s of the variable using the Zero Product Property. Finally, it is a must to do checking of the solution.
$\bf{\text{Solution Details:}}$
Using the properties of equality to isolate the radical, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{6x+7}=x-7+9
\\\\
\sqrt{6x+7}=x+2
.\end{array}
Squaring both sides and then using the special product on squaring binomials 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}
6x+7=(x+2)^2
\\\\
6x+7=[(x)^2+2(x)(2)+(2)^2]
\\\\
6x+7=x^2+4x+4
\\\\
-x^2+(6x-4x)+(7-4)=0
\\\\
-x^2+2x+3=0
\\\\
-1(-x^2+2x+3)=(0)(-1)
\\\\
x^2-2x-3=0
.\end{array}
In the trinomial expression above, the value of $ac$ is $
1(-3)=-3
$ and the value of $b$ is $
-2
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
1,-3
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x^2+x-3x-3=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^2+x)-(3x+3)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(x+1)-3(x+1)=0
.\end{array}
Factoring the $GCF=
(x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x+1)(x-3)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x+1=0
\\\\\text{OR}\\\\
x-3=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x+1=0
\\\\
x=-1
\\\\\text{OR}\\\\
x-3=0
\\\\
x=3
.\end{array}
Upon checking, $
x=\left\{ -1,3 \right\}
$ satisfy the original equation.