Answer
$x=\left\{ \dfrac{-2-i\sqrt{2}}{3},\dfrac{-2+i\sqrt{2}}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
x(3x+4)=-2
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x(3x)+x(4)=-2
\\\\
3x^2+4x=-2
.\end{array}
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
3x^2+4x+2=0
.\end{array}
The quadratic equation above has $a=
3
, b=
4
, c=
2
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm\sqrt{4^2-4(3)(2)}}{2(3)}
\\\\
x=\dfrac{-4\pm\sqrt{16-24}}{6}
\\\\
x=\dfrac{-4\pm\sqrt{-8}}{6}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm\sqrt{-1}\cdot\sqrt{8}}{6}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm i\sqrt{8}}{6}
.\end{array}
Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm i\sqrt{4\cdot2}}{6}
\\\\
x=\dfrac{-4\pm i\sqrt{(2)^2\cdot2}}{6}
\\\\
x=\dfrac{-4\pm i(2)\sqrt{2}}{6}
\\\\
x=\dfrac{-4\pm 2i\sqrt{2}}{6}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel2(-2)\pm \cancel2(1)i\sqrt{2}}{\cancel2(3)}
\\\\
x=\dfrac{-2\pm i\sqrt{2}}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-2-i\sqrt{2}}{3}
\\\\\text{OR}\\\\
x=\dfrac{-2+i\sqrt{2}}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{-2-i\sqrt{2}}{3},\dfrac{-2+i\sqrt{2}}{3} \right\}
.$
