## College Algebra (11th Edition)

$x=\left\{ \dfrac{-1-i\sqrt{14}}{3}, \dfrac{-1+i\sqrt{14}}{3} \right\}$
$\bf{\text{Solution Outline:}}$ 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 $x=\left\{ \dfrac{-1-i\sqrt{14}}{3}, \dfrac{-1+i\sqrt{14}}{3} \right\} .$