Answer
$\left\{-\dfrac{3}{10}-\dfrac{1}{10}i,\dfrac{3}{10}+\dfrac{1}{10}i \right\}$.
Work Step by Step
The given equation has $a=10,b=6$ and $c=1$.
The discriminant is
$=b^2-4ac$
$=(6)^2-4(10)(1)$
$=36-40$
$=-4$
Use Quadratic formula.
$x=\dfrac{-b\pm \sqrt {b^2-4ac}}{2a}$
Substitute the values of $a, b, c, $ and discriminant to obtain:
$x=\dfrac{-(6)\pm \sqrt {-4}}{2(10)}$
Simplify.
$x=\dfrac{-6\pm \sqrt{4(-1)}}{20}$
$x=\dfrac{-6\pm 2i}{20}$
Factor out $2$.
$x=\dfrac{2(-3\pm i)}{20}$
Cancel the common factor $2$.
$x=\dfrac{-3\pm i}{10}$
The equation has two solutions: $-\frac{3}{10}-\frac{i}{10}$ and $-\frac{3}{10}+\frac{i}{10}$.
Check:
For $x=-\frac{3}{10}-\frac{i}{10}$.
$=10(-\frac{3}{10}-\frac{i}{10})^2+6(-\frac{3}{10}-\frac{i}{10})+1$
$=10(\frac{9}{100}+2\cdot \frac{3i}{100}+\frac{i^2}{100})-\frac{9}{5}-\frac{3i}{5}+1$
$=\frac{9}{10}+\frac{3i}{5}+\frac{i^2}{10}-\frac{9}{5}-\frac{3i}{5}+1$
$=\frac{1}{10}+\frac{i^2}{10}$
$=\frac{1}{10}-\frac{1}{10}$
$=0$
For $x=-\frac{3}{10}+\frac{i}{10}$.
$=10(-\frac{3}{10}+\frac{i}{10})^2+6(-\frac{3}{10}+\frac{i}{10})+1$
$=10(\frac{9}{100}-2\cdot \frac{3i}{100}+\frac{i^2}{100})-\frac{9}{5}+\frac{3i}{5}+1$
$=\frac{9}{10}-\frac{3i}{5}+\frac{i^2}{10}-\frac{9}{5}+\frac{3i}{5}+1$
$=\frac{1}{10}+\frac{i^2}{10}$
$=\frac{1}{10}-\frac{1}{10}$
$=0$
Hence, the solution set is $\left\{-\dfrac{3}{10}-\dfrac{1}{10}i,\dfrac{3}{10}+\dfrac{1}{10}i \right\}$.
