Answer
$x=\left\{ \dfrac{-3-i\sqrt{7}}{4},\dfrac{-3+i\sqrt{7}}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
z(2z+3)=-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}
z(2z)+z(3)=-2
\\\\
2z^2+3z=-2
.\end{array}
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
2z^2+3z+2=0
.\end{array}
The quadratic equation above has $a=
2
, b=
3
, c=
2
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-3\pm\sqrt{3^2-4(2)(2)}}{2(2)}
\\\\
x=\dfrac{-3\pm\sqrt{9-16}}{4}
\\\\
x=\dfrac{-3\pm\sqrt{-7}}{4}
.\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{-3\pm\sqrt{-1}\cdot\sqrt{7}}{4}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{-3\pm i\sqrt{7}}{4}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-3-i\sqrt{7}}{4}
\\\\\text{OR}\\\\
x=\dfrac{-3+i\sqrt{7}}{4}
.\end{array}
Hence, $
x=\left\{ \dfrac{-3-i\sqrt{7}}{4},\dfrac{-3+i\sqrt{7}}{4} \right\}
.$
