Answer
$x=\left\{ \dfrac{-3-\sqrt{21}}{3},\dfrac{-3+\sqrt{21}}{3}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
-3x(x+2)=-4
,$ 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}
-3x(x)-3x(2)=-4
\\\\
-3x^2-6x=-4
.\end{array}
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-3x^2-6x+4=0
\\\\
-1(-3x^2-6x+4)=-1(0)
\\\\
3x^2+6x-4=0
.\end{array}
The quadratic equation above has $a=
3
, b=
6
, c=
-4
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-6\pm\sqrt{6^2-4(3)(-4)}}{2(3)}
\\\\
x=\dfrac{-6\pm\sqrt{36+48}}{6}
\\\\
x=\dfrac{-6\pm\sqrt{84}}{6}
.\end{array}
Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to
\begin{array}{l}\require{cancel}
x=\dfrac{-6\pm\sqrt{4\cdot21}}{6}
\\\\
x=\dfrac{-6\pm\sqrt{(2)^2\cdot21}}{6}
\\\\
x=\dfrac{-6\pm2\sqrt{21}}{6}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel{2}(-3)\pm\cancel{2}(1)\sqrt{21}}{\cancel{2}(3)}
\\\\
x=\dfrac{-3\pm\sqrt{21}}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-3-\sqrt{21}}{3}
\\\\\text{OR}\\\\
x=\dfrac{-3+\sqrt{21}}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{-3-\sqrt{21}}{3},\dfrac{-3+\sqrt{21}}{3}
\right\}
.$
