#### Answer

$x=\left\{ -3,1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt[4]{x^2+2x}=\sqrt[4]{3}
,$ raise both sides to the fourth power. 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. Checking of the solution is a must since both sides were raised to an even power.
$\bf{\text{Solution Details:}}$
Raising both sides to the fourth power and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^2+2x=3
\\\\
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\{ -3,1 \right\}
$ satisfy the original equation.