We are given a quadratic equation $2x^2 + 4x - 6 = 0$ and we need to find the values of $x$ that satisfy this equation.

1. Problem Description

We are given a quadratic equation $2x^2 + 4x - 6 = 0$ and we need to find the values of $x$ that satisfy this equation.

2. Solution Steps

First, we can simplify the equation by dividing all terms by 2:

$x^2 + 2x - 3 = 0$

Next, we can try to factor the quadratic expression. We are looking for two numbers that multiply to -3 and add up to

2. These numbers are 3 and -

1. Thus, we can factor the equation as:

$(x + 3)(x - 1) = 0$

Now, we can set each factor equal to zero and solve for $x$:

$x + 3 = 0$ or $x - 1 = 0$

Solving for $x$ in each case gives:

$x = -3$ or $x = 1$

Therefore, the solutions to the quadratic equation are $x = -3$ and $x = 1$.

Alternatively, we can use the quadratic formula to solve for $x$. The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $x^2 + 2x - 3 = 0$, we have $a = 1$, $b = 2$, and $c = -3$. Plugging these values into the quadratic formula, we get:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)}$

$x = \frac{-2 \pm \sqrt{4 + 12}}{2}$

$x = \frac{-2 \pm \sqrt{16}}{2}$

$x = \frac{-2 \pm 4}{2}$

$x = \frac{-2 + 4}{2}$ or $x = \frac{-2 - 4}{2}$

$x = \frac{2}{2}$ or $x = \frac{-6}{2}$

$x = 1$ or $x = -3$

3. Final Answer

The solutions to the equation are $x = -3$ and $x = 1$.