Question - Solving a Quadratic Equation

We have the quadratic equation \(2x^2 - 4x - 6 = 0\).

To solve for \(x\), we can use the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a = 2\), \(b = -4\), and \(c = -6\).

First, calculate the discriminant (\(b^2 - 4ac\)): \((-4)^2 - 4 \cdot 2 \cdot (-6) = 16 + 48 = 64\).

Since the discriminant is positive, there are two real solutions.

Now compute the two solutions using the quadratic formula:

\(x = \frac{{-(-4) \pm \sqrt{64}}}{2 \cdot 2} = \frac{{4 \pm 8}}{4}\).

So, the two solutions are:

\(x = \frac{{4 + 8}}{4} = \frac{{12}}{4} = 3\)

and

\(x = \frac{{4 - 8}}{4} = \frac{{-4}}{4} = -1\).

Thus, the solutions to the equation \(2x^2 - 4x - 6 = 0\) are \(x = 3\) and \(x = -1\).