1. **State the problem:** Solve the quadratic equation $2x^2 + 6x = 2$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$2x^2 + 6x - 2 = 0$$ 3. **Simplify the equation:** Divide every term by 2 to simplify: $$\cancel{2}x^2 + \cancel{2} \cdot 3x - \cancel{2} = 0 \implies x^2 + 3x - 1 = 0$$ 4. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, the solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=3$, and $c=-1$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-1) = 9 + 4 = 13$$ 6. **Find the roots:** $$x = \frac{-3 \pm \sqrt{13}}{2 \cdot 1} = \frac{-3 \pm \sqrt{13}}{2}$$ 7. **Final answer:** The solutions to the equation are $$x = \frac{-3 + \sqrt{13}}{2} \quad \text{and} \quad x = \frac{-3 - \sqrt{13}}{2}$$