1. **State the problem:** Solve the equation $2x - 2 = \sqrt{3x^2 + 13}$ for $x$. 2. **Formula and rules:** To solve equations involving square roots, we isolate the square root term and then square both sides to eliminate the root. Remember to check for extraneous solutions after squaring. 3. **Isolate the square root:** The equation is already isolated: $$2x - 2 = \sqrt{3x^2 + 13}$$ 4. **Square both sides:** $$ (2x - 2)^2 = (\sqrt{3x^2 + 13})^2 $$ $$ (2x - 2)^2 = 3x^2 + 13 $$ 5. **Expand the left side:** $$ (2x - 2)^2 = (2x)^2 - 2 \times 2x \times 2 + 2^2 = 4x^2 - 8x + 4 $$ 6. **Set up the quadratic equation:** $$ 4x^2 - 8x + 4 = 3x^2 + 13 $$ 7. **Bring all terms to one side:** $$ 4x^2 - 8x + 4 - 3x^2 - 13 = 0 $$ $$ (4x^2 - 3x^2) - 8x + (4 - 13) = 0 $$ $$ x^2 - 8x - 9 = 0 $$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=1$, $b=-8$, $c=-9$. Calculate the discriminant: $$ \Delta = (-8)^2 - 4 \times 1 \times (-9) = 64 + 36 = 100 $$ Calculate the roots: $$ x = \frac{-(-8) \pm \sqrt{100}}{2 \times 1} = \frac{8 \pm 10}{2} $$ Two solutions: $$ x_1 = \frac{8 + 10}{2} = \frac{18}{2} = 9 $$ $$ x_2 = \frac{8 - 10}{2} = \frac{-2}{2} = -1 $$ 9. **Check for extraneous solutions:** Substitute $x=9$ into the original equation: $$ 2(9) - 2 = 18 - 2 = 16 $$ $$ \sqrt{3(9)^2 + 13} = \sqrt{3 \times 81 + 13} = \sqrt{243 + 13} = \sqrt{256} = 16 $$ Both sides equal 16, so $x=9$ is valid. Substitute $x=-1$: $$ 2(-1) - 2 = -2 - 2 = -4 $$ $$ \sqrt{3(-1)^2 + 13} = \sqrt{3 \times 1 + 13} = \sqrt{16} = 4 $$ Left side is $-4$, right side is $4$, not equal, so $x=-1$ is extraneous. **Final answer:** $$ \boxed{9} $$