For what values of x is the equation `2x^(2)=4x+3` true?

Options

A. `(2pmsqrt(10))/(2)`
B. `(2pm5sqrt(2))/(2)`
C. `1pmsqrt(5)`
D. `1pmsqrt(10)`

Explanation

To solve the equation \(2x^2 = 4x + 3\), we will follow these steps:

### Step 1: Rearrange the equation
First, we need to rearrange the equation into the standard quadratic form \(ax^2 + bx + c = 0\).

Starting with the original equation:
\[ 2x^2 = 4x + 3 \]

Subtract \(4x\) and \(3\) from both sides:
\[ 2x^2 - 4x - 3 = 0 \]

### Step 2: Identify coefficients
Now, we identify the coefficients \(a\), \(b\), and \(c\) from the equation \(2x^2 - 4x - 3 = 0\):
- \(a = 2\)
- \(b = -4\)
- \(c = -3\)

### Step 3: Use the quadratic formula
The quadratic formula to find the roots of the equation \(ax^2 + bx + c = 0\) is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 4: Calculate the discriminant
First, we calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = (-4)^2 = 16 \]
\[ 4ac = 4 \cdot 2 \cdot (-3) = -24 \]
Now, substituting these values into the discriminant:
\[ b^2 - 4ac = 16 - (-24) = 16 + 24 = 40 \]

### Step 5: Substitute into the quadratic formula
Now we substitute \(a\), \(b\), and the discriminant into the quadratic formula:
\[ x = \frac{-(-4) \pm \sqrt{40}}{2 \cdot 2} \]
This simplifies to:
\[ x = \frac{4 \pm \sqrt{40}}{4} \]

### Step 6: Simplify the square root
We can simplify \(\sqrt{40}\):
\[ \sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10} \]
Now substituting this back into the equation:
\[ x = \frac{4 \pm 2\sqrt{10}}{4} \]
This can be simplified further:
\[ x = 1 \pm \frac{\sqrt{10}}{2} \]

### Step 7: Final values of x
Thus, the values of \(x\) are:
\[ x = 1 + \frac{\sqrt{10}}{2} \quad \text{and} \quad x = 1 - \frac{\sqrt{10}}{2} \]