Find the real solutions to the equation. $$ \frac{6 x^{2}+5}{18 x^{2}-6 x-3}=\frac{1}{3} $$ Real solution(s): $$ \{\square\} $$

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Solve the equation $$ \frac{6x^{2}+5}{18x^{2}-6x-3}=\frac{1}{3} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{6x^{2}+5}{18x^{2}-6x-3}=\frac{1}{3}$$
- step1: Find the domain:
  $$\frac{6x^{2}+5}{18x^{2}-6x-3}=\frac{1}{3},x \in \left(-\infty,\frac{-\sqrt{7}+1}{6}ight)\cup \left(\frac{-\sqrt{7}+1}{6},\frac{\sqrt{7}+1}{6}ight)\cup \left(\frac{\sqrt{7}+1}{6},+\inftyight)$$
- step2: Cross multiply:
  $$\left(6x^{2}+5ight)	imes 3=18x^{2}-6x-3$$
- step3: Simplify the equation:
  $$3\left(6x^{2}+5ight)=18x^{2}-6x-3$$
- step4: Rewrite the expression:
  $$3\left(6x^{2}+5ight)=3\left(6x^{2}-2x-1ight)$$
- step5: Evaluate:
  $$6x^{2}+5=6x^{2}-2x-1$$
- step6: Cancel equal terms:
  $$5=-2x-1$$
- step7: Swap the sides:
  $$-2x-1=5$$
- step8: Move the constant to the right side:
  $$-2x=5+1$$
- step9: Add the numbers:
  $$-2x=6$$
- step10: Change the signs:
  $$2x=-6$$
- step11: Divide both sides:
  $$\frac{2x}{2}=\frac{-6}{2}$$
- step12: Divide the numbers:
  $$x=-3$$
- step13: Check if the solution is in the defined range:
  $$x=-3,x \in \left(-\infty,\frac{-\sqrt{7}+1}{6}ight)\cup \left(\frac{-\sqrt{7}+1}{6},\frac{\sqrt{7}+1}{6}ight)\cup \left(\frac{\sqrt{7}+1}{6},+\inftyight)$$
- step14: Find the intersection:
  $$x=-3$$
The real solution to the equation is $$ x = -3 $$.
The real solution is $$ x = -3 $$.

Find the real solutions to the equation. \[ \frac{6 x^{2}+5}{18 x^{2}-6 x-3}=\frac{1}{3} \] Real solution(s): \( \{\square\} \)

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