#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
3x^2+3=-10x
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
3x^2+10x+3=0
\\\\
(3x+1)(x+3)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
3x+1=0
\text{ OR }
x+3=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
3x+1=0
\\\\
3x=-1
\\\\
x=-\dfrac{1}{3}
\\\\\text{ OR }\\\\
x+3=0
\\\\
x=-3
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{1}{3},-3 \right\}
.$
$\bf{\text{Supplementary Solution:}}$
To factor the expression, $
3x^2+10x+3
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
In the expression above the value of $ac$ is $
3(3)=9
$ and the value of $b$ is $
10
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,9\}, \{3,3\},
\{-1,-9\}, \{-3,-3\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,9
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3x^2+1x+9x+3
.\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}
(3x^2+1x)+(9x+3)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(3x+1)+3(3x+1)
.\end{array}
Factoring the $GCF=
(3x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3x+1)(x+3)
.\end{array}