Find the real solutions, if any, using the quadratic for $$ 2 y^{2}-y+8=0 $$

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Solve the equation $$ 2y^{2}-y+8=0 $$.
Solve the equation(The complex numbers system) by following steps:
- step0: Solve using the quadratic formula in the complex numbers system:
  $$2y^{2}-y+8=0$$
- step1: Solve using the quadratic formula:
  $$y=\frac{1\pm \sqrt{\left(-1ight)^{2}-4	imes 2	imes 8}}{2	imes 2}$$
- step2: Simplify the expression:
  $$y=\frac{1\pm \sqrt{\left(-1ight)^{2}-4	imes 2	imes 8}}{4}$$
- step3: Simplify the expression:
  $$y=\frac{1\pm \sqrt{-63}}{4}$$
- step4: Simplify the expression:
  $$y=\frac{1\pm 3\sqrt{7}	imes i}{4}$$
- step5: Separate into possible cases:
  $$\begin{align}&y=\frac{1+3\sqrt{7}	imes i}{4}\&y=\frac{1-3\sqrt{7}	imes i}{4}\end{align}$$
- step6: Simplify the expression:
  $$\begin{align}&y=\frac{1}{4}+\frac{3\sqrt{7}}{4}i\&y=\frac{1-3\sqrt{7}	imes i}{4}\end{align}$$
- step7: Simplify the expression:
  $$\begin{align}&y=\frac{1}{4}+\frac{3\sqrt{7}}{4}i\&y=\frac{1}{4}-\frac{3\sqrt{7}}{4}i\end{align}$$
- step8: Rewrite:
  $$y_{1}=\frac{1}{4}-\frac{3\sqrt{7}}{4}i,y_{2}=\frac{1}{4}+\frac{3\sqrt{7}}{4}i$$
- step9: Remove the complex number(s):
  $$	extrm{No real solution}$$
The given quadratic equation $$2y^{2}-y+8=0$$ does not have any real solutions.
There are no real solutions to the equation $$2y^{2}-y+8=0$$.

Find the real solutions, if any, using the quadratic for

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