#### Answer

$x=5$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{4x+5}-6=2x-11
,$ 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{4x+5}=2x-11+6
\\\\
\sqrt{4x+5}=2x-5
.\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}
4x+5=(2x-5)^2
\\\\
4x+5=[(2x)^2-2(2x)(5)+(5)^2]
\\\\
4x+5=4x^2-20x+25
\\\\
-4x^2+(4x+20x)+(5-25)=0
\\\\
-4x^2+24x-20=0
\\\\
\dfrac{-4x^2+24x-20}{-4}=\dfrac{0}{-4}
\\\\
x^2-6x+5=0
.\end{array}
In the trinomial expression above, the value of $ac$ is $
1(5)=5
$ and the value of $b$ is $
-6
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-1,-5
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x^2-x-5x+5=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)-(5x-5)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(x-1)-5(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-5)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x-1=0
\\\\\text{OR}\\\\
x-5=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-1=0
\\\\
x=1
\\\\\text{OR}\\\\
x-5=0
\\\\
x=5
.\end{array}
Upon checking, only $
x=5
$ satisfies the original equation.