Answer
$x=\left\{ \dfrac{-1-i\sqrt{14}}{3}, \dfrac{-1+i\sqrt{14}}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
-x(3x+2)=5
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula to solve for $x.$
$\bf{\text{Solution Details:}}$
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
-x(3x)-x(2)=5
\\\\
-3x^2-2x=5
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
-3x^2-2x-5=0
\\\\
-1(-3x^2-2x-5)=-1(0)
\\\\
3x^2+2x+5=0
.\end{array}
In the equation above, $a=
3
,$ $b=
2
,$ and $c=
5
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-2\pm\sqrt{2^2-4(3)(5)}}{2(3)}
\\\\
x=\dfrac{-2\pm\sqrt{4-60}}{6}
\\\\
x=\dfrac{-2\pm\sqrt{-56}}{6}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{-2\pm\sqrt{-1}\cdot\sqrt{56}}{6}
\\\\
x=\dfrac{-2\pm i\cdot\sqrt{4\cdot14}}{6}
\\\\
x=\dfrac{-2\pm i\cdot\sqrt{(2)^2\cdot14}}{6}
\\\\
x=\dfrac{-2\pm i\cdot2\sqrt{14}}{6}
\\\\
x=\dfrac{-2\pm 2i\sqrt{14}}{6}
\\\\
x=\dfrac{\cancel2^{-1}\pm \cancel2^1i\sqrt{14}}{\cancel2^3}
\\\\
x=\dfrac{-1\pm i\sqrt{14}}{3}
.\end{array}
The solutions are $
x=\left\{ \dfrac{-1-i\sqrt{14}}{3}, \dfrac{-1+i\sqrt{14}}{3} \right\}
.$
