Solve the Quadratic Equation: x² + 3x - 3 = x

To solve the quadratic equation $x^2 + 3x - 3 = x$, we first rearrange it to standard form.

Step 1: Move all terms to one side of the equation:
$x^2 + 3x - 3 - x = 0$
This simplifies to:
$x^2 + 2x - 3 = 0$

Step 2: Identify the coefficients in the standard form $ax^2 + bx + c = 0$:
Here, $a = 1$, $b = 2$, and $c = -3$.

Step 3: Use the quadratic formula:
$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$

Plug in the values for $a$, $b$, and $c$:
$x = \frac{{-2 \pm \sqrt{{(2)^2 - 4 \times 1 \times (-3)}}}}{2 \times 1}$

Simplify under the square root:
$x = \frac{{-2 \pm \sqrt{{4 + 12}}}}{2} = \frac{{-2 \pm \sqrt{16}}}{2}$

Simplify further:
$x = \frac{{-2 \pm 4}}{2}$

This results in two solutions:
For the positive square root:
$x = \frac{{-2 + 4}}{2} = \frac{2}{2} = 1$

For the negative square root:
$x = \frac{{-2 - 4}}{2} = \frac{-6}{2} = -3$

Therefore, the solutions are $x_1 = 1$, $x_2 = -3$.

Since these solutions match choice $x_1 = 1$, $x_2 = -3$, we verify accuracy against the answer choices.

The correct solution to the equation is $x_1 = 1$ and $x_2 = -3$.