#### Answer

$x=3$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{4x+13}=2x-1
,$ square both sides and 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.
$\bf{\text{Solution Details:}}$
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+13=(2x-1)^2
\\\\
4x+13=[(2x)^2-2(2x)(1)+(1)^2]
\\\\
4x+13=4x^2-4x+1
\\\\
-4x^2+(4x+4x)+(13-1)=0
\\\\
-4x^2+8x+12=0
\\\\
\dfrac{-4x^2+8x+12}{-4}=\dfrac{0}{-4}
\\\\
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, only $
x=3
$ satisfies the original equation.