Question

solved algebra question involving quadratic equations with a variable 'z'

Answer 1

Sure! Let's work through a quadratic equation involving the variable \( z \).

Consider the quadratic equation:

\[
2z^2 - 4z - 6 = 0
\]

### Step 1: Identify coefficients
In the standard form of a quadratic equation \( az^2 + bz + c = 0 \), we have:
- \( a = 2 \)
- \( b = -4 \)
- \( c = -6 \)

### Step 2: Use the quadratic formula
The solutions for \( z \) can be found using the quadratic formula:

\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

### Step 3: Calculate the discriminant
First, we calculate the discriminant \( D = b^2 - 4ac \):

\[
D = (-4)^2 - 4 \cdot 2 \cdot (-6)
\]
\[
D = 16 + 48 = 64
\]

### Step 4: Substitute values into the formula
Now, substituting the values back into the quadratic formula:

\[
z = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 2}
\]
\[
z = \frac{4 \pm 8}{4}
\]

### Step 5: Solve for the two values of \( z \)
Now, we calculate the two potential solutions:

1. \( z_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3 \)
2. \( z_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1 \)

### Step 6: Final solutions
Thus, the solutions for the equation \( 2z^2 - 4z - 6 = 0 \) are:

\[
z = 3 \quad \text{and} \quad z = -1
\]

If you have a different equation in mind or need further assistance, feel free to ask!