#### Answer

$x=\left\{ -6,-5 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
3x^2+9x+30=2x^2-2x
,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
Using the properties of equality and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(3x^2-2x^2)+(9x+2x)+30=0
\\\\
x^2+11x+30=0
.\end{array}
To factor the trinomial expression above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ In the expression above, the value of $c$ is $
30
$ and the value of $b$ is $
11
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,30 \}, \{ 2,15 \}, \{ 3,10 \}, \{ 5,6 \},
\\
\{ -1,-30 \}, \{ -2,-15 \}, \{ -3,-10 \}, \{ -5,-6 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
5,6
\}.$ Hence, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x+5)(x+6)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
x+5=0
\text{ OR }
x+6=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
x+5=0
\\\\
x=-5
\\\\\text{ OR }\\\\
x+6=0
\\\\
x=-6
.\end{array}
Hence, the solutions are $
x=\left\{ -6,-5 \right\}
.$