$$ 6 x^{2}+6 x+1=0 $$ determine if it is ane repeated solution, two different real solutions or two non-real solutions.

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Solve the equation $$ 6x^{2}+6x+1=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve using the quadratic formula:
  $$6x^{2}+6x+1=0$$
- step1: Solve using the quadratic formula:
  $$x=\frac{-6\pm \sqrt{6^{2}-4	imes 6}}{2	imes 6}$$
- step2: Simplify the expression:
  $$x=\frac{-6\pm \sqrt{6^{2}-4	imes 6}}{12}$$
- step3: Simplify the expression:
  $$x=\frac{-6\pm \sqrt{12}}{12}$$
- step4: Simplify the expression:
  $$x=\frac{-6\pm 2\sqrt{3}}{12}$$
- step5: Separate into possible cases:
  $$\begin{align}&x=\frac{-6+2\sqrt{3}}{12}\&x=\frac{-6-2\sqrt{3}}{12}\end{align}$$
- step6: Simplify the expression:
  $$\begin{align}&x=\frac{-3+\sqrt{3}}{6}\&x=\frac{-6-2\sqrt{3}}{12}\end{align}$$
- step7: Simplify the expression:
  $$\begin{align}&x=\frac{-3+\sqrt{3}}{6}\&x=-\frac{3+\sqrt{3}}{6}\end{align}$$
- step8: Rewrite:
  $$x_{1}=-\frac{3+\sqrt{3}}{6},x_{2}=\frac{-3+\sqrt{3}}{6}$$
The solutions to the equation $$6x^{2}+6x+1=0$$ are $$x_{1}=-\frac{3+\sqrt{3}}{6}$$ and $$x_{2}=\frac{-3+\sqrt{3}}{6}$$.

Since the solutions are different real numbers, the equation has two different real solutions.
The equation has two different real solutions.

\[ 6 x^{2}+6 x+1=0 \] determine if it is ane repeated solution, two different real solutions or two non-real solutions.

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