(1) $$ \sqrt{2 x^{2}-x-1}+1=x $$ (2) $$ \sqrt{x+2}-x+4=0 $$ (3) $$ 6=\sqrt{x+12}-x $$ (4) $$ x+\sqrt{-4 x-3}=0 $$

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Solve the equation $$ \sqrt{2x^{2}-x-1}+1=x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sqrt{2x^{2}-x-1}+1=x$$
- step1: Find the domain:
  $$\sqrt{2x^{2}-x-1}+1=x ,x \in \left(-\infty,-\frac{1}{2}ight]\cup \left[1,+\inftyight)$$
- step2: Move the constant to the right-hand side:
  $$\sqrt{2x^{2}-x-1}=x-1$$
- step3: Evaluate:
  $$\sqrt{2x^{2}-x-1}=x-1,x-1\geq 0$$
- step4: Evaluate:
  $$\sqrt{2x^{2}-x-1}=x-1,x\geq 1$$
- step5: Solve the equation:
  $$\begin{align}&x=1\&x=-2\end{align},x\geq 1$$
- step6: Find the intersection:
  $$x=1$$
- step7: Check if the solution is in the defined range:
  $$x=1,x \in \left(-\infty,-\frac{1}{2}ight]\cup \left[1,+\inftyight)$$
- step8: Find the intersection:
  $$x=1$$
- step9: Check the solution:
  $$x=1$$
Solve the equation $$ \sqrt{x+2}-x+4=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sqrt{x+2}-x+4=0$$
- step1: Find the domain:
  $$\sqrt{x+2}-x+4=0,x\geq -2$$
- step2: Move the expression to the right-hand side:
  $$\sqrt{x+2}=x-4$$
- step3: Evaluate:
  $$\sqrt{x+2}=x-4,x-4\geq 0$$
- step4: Evaluate:
  $$\sqrt{x+2}=x-4,x\geq 4$$
- step5: Solve the equation:
  $$\begin{align}&x=7\&x=2\end{align},x\geq 4$$
- step6: Find the intersection:
  $$x=7$$
- step7: Check if the solution is in the defined range:
  $$x=7,x\geq -2$$
- step8: Find the intersection:
  $$x=7$$
- step9: Check the solution:
  $$x=7$$
Solve the equation $$ 6=\sqrt{x+12}-x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$6=\sqrt{x+12}-x$$
- step1: Find the domain:
  $$6=\sqrt{x+12}-x,x\geq -12$$
- step2: Swap the sides:
  $$\sqrt{x+12}-x=6$$
- step3: Move the variable to the right-hand side:
  $$\sqrt{x+12}=6+x$$
- step4: Evaluate:
  $$\sqrt{x+12}=6+x,6+x\geq 0$$
- step5: Evaluate:
  $$\sqrt{x+12}=6+x,x\geq -6$$
- step6: Solve the equation:
  $$\begin{align}&x=-3\&x=-8\end{align},x\geq -6$$
- step7: Find the intersection:
  $$x=-3$$
- step8: Check if the solution is in the defined range:
  $$x=-3,x\geq -12$$
- step9: Find the intersection:
  $$x=-3$$
- step10: Check the solution:
  $$x=-3$$
Solve the equation $$ x+\sqrt{-4x-3}=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x+\sqrt{-4x-3}=0$$
- step1: Find the domain:
  $$x+\sqrt{-4x-3}=0,x\leq -\frac{3}{4}$$
- step2: Move the variable to the right-hand side:
  $$\sqrt{-4x-3}=-x$$
- step3: Evaluate:
  $$\sqrt{-4x-3}=-x,-x\geq 0$$
- step4: Evaluate:
  $$\sqrt{-4x-3}=-x,x\leq 0$$
- step5: Solve the equation:
  $$\begin{align}&x=-1\&x=-3\end{align},x\leq 0$$
- step6: Find the intersection:
  $$\begin{align}&x=-1\&x=-3\end{align}$$
- step7: Check if the solution is in the defined range:
  $$\begin{align}&x=-1\&x=-3\end{align},x\leq -\frac{3}{4}$$
- step8: Find the intersection:
  $$\begin{align}&x=-1\&x=-3\end{align}$$
- step9: Check the solution:
  $$\begin{align}&x=-1\&x=-3\end{align}$$
- step10: Rewrite:
  $$x_{1}=-3,x_{2}=-1$$
The solutions to the given equations are:
1. $$ x = 1 $$
2. $$ x = 7 $$
3. $$ x = -3 $$
4. $$ x = -3 $$ or $$ x = -1 $$
The solutions are: 1. $$ x = 1 $$ 2. $$ x = 7 $$ 3. $$ x = -3 $$ 4. $$ x = -3 $$ or $$ x = -1 $$

(1) \( \sqrt{2 x^{2}-x-1}+1=x \) (2) \( \sqrt{x+2}-x+4=0 \) (3) \( 6=\sqrt{x+12}-x \) (4) \( x+\sqrt{-4 x-3}=0 \)

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