Answer
$x=\left\{ \dfrac{3-i\sqrt{2}}{2},\dfrac{3+i\sqrt{2}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
-4x^2=-12x+11
,$ express the equation in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
-4x^2+12x-11=0
\\\\
\dfrac{-4x^2+12x-11}{-4}=\dfrac{0}{-4}
\\\\
x^2-3x+\dfrac{11}{4}=0
.\end{array}
In the equation above, $a=
1
,$ $b=
-3
,$ and $c=
\dfrac{11}{4}
.$ 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{-(-3)\pm\sqrt{(-3)^2-4(1)\left(\dfrac{11}{4}\right)}}{2(1)}
\\\\
x=\dfrac{3\pm\sqrt{9-11}}{2}
\\\\
x=\dfrac{3\pm\sqrt{-2}}{2}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and that $i=\sqrt{-1},$ the expression above is equivalent to\begin{array}{l}\require{cancel}
x=\dfrac{3\pm\sqrt{-1}\cdot\sqrt{2}}{2}
\\\\=
x=\dfrac{3\pm i\sqrt{2}}{2}
.\end{array}
The solutions are $
x=\left\{ \dfrac{3-i\sqrt{2}}{2},\dfrac{3+i\sqrt{2}}{2} \right\}
.$
