Answer
$\left\{\dfrac{1}{5}-\dfrac{2}{5}i,\dfrac{1}{5}+\dfrac{2}{5}i \right\}$
Work Step by Step
Subtract $2x$ from both sides.
$5x^2+1-2x=2x-2x$
Simplify.
$5x^2-2x+1=0$
The quadratic equation above has $a=5,b=-2$ and $c=1$.
The discriminant is
$=b^2-4ac$
$=(-2)^2-4(5)(1)$
$=4-20$
$=-16$
Use Quadratic formula.
$x=\dfrac{-b\pm \sqrt {b^2-4ac}}{2a}$
Substitute the values of $a, b, c, $ and discriminant to obtain:
$x=\dfrac{-(-2)\pm \sqrt {-16}}{2(5)}$
Simplify.
$x=\dfrac{2\pm \sqrt{16(-1)}}{10}$
$x=\dfrac{2\pm 4i}{10}$
Factor out $2$.
$x=\dfrac{2(1\pm 2i)}{10}$
Cancel the common factor $2$.
$x=\dfrac{1\pm 2i}{5}$
The equation has two solutions: $\frac{1}{5}-\frac{2i}{5}$ and $\frac{1}{5}+\frac{2i}{5}$.
Check:
For $x=\frac{1}{5}-\frac{2i}{5}$.
$=5(\frac{1}{5}-\frac{2i}{5})^2-2(\frac{1}{5}-\frac{2i}{5})+1$
$=5(\frac{1}{25}-2\cdot \frac{2i}{25}+\frac{4i^2}{25})-\frac{2}{5}+\frac{4i}{5}+1$
$=\frac{1}{5}-\frac{4i}{5}+\frac{4i^2}{5}-\frac{2}{5}+\frac{4i}{5}+1$
$=\frac{4}{5}+\frac{4i^2}{5}$
$=\frac{4}{5}-\frac{4}{5}$
$=0$
For $x=\frac{1}{5}+\frac{2i}{5}$.
$=5(\frac{1}{5}+\frac{2i}{5})^2-2(\frac{1}{5}+\frac{2i}{5})+1$
$=5(\frac{1}{25}+2\cdot \frac{2i}{25}+\frac{4i^2}{25})-\frac{2}{5}-\frac{4i}{5}+1$
$=\frac{1}{5}+\frac{4i}{5}+\frac{4i^2}{5}-\frac{2}{5}-\frac{4i}{5}+1$
$=\frac{4}{5}+\frac{4i^2}{5}$
$=\frac{4}{5}-\frac{4}{5}$
$=0$
Hence, the solution set is $\left\{\dfrac{1}{5}-\dfrac{2}{5}i,\dfrac{1}{5}+\dfrac{2}{5}i \right\}$.
