Find `x` if `(x+1)(2x-3)=4`.

Answer

To solve the equation \((x+1)(2x-3)=4\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:

\[ (x + 1)(2x - 3) = x \cdot 2x + x \cdot (-3) + 1 \cdot 2x + 1 \cdot (-3) \]

This simplifies to:

\[ 2x^2 - 3x + 2x - 3 = 2x^2 - x - 3 \]

### Step 2: Set the equation to zero
Now, we set the equation equal to 4:

\[ 2x^2 - x - 3 = 4 \]

Next, we subtract 4 from both sides to set the equation to zero:

\[ 2x^2 - x - 3 - 4 = 0 \]

This simplifies to:

\[ 2x^2 - x - 7 = 0 \]

### Step 3: Identify coefficients
In the quadratic equation \(ax^2 + bx + c = 0\), we identify:

- \(a = 2\)
- \(b = -1\)
- \(c = -7\)

### Step 4: Apply the quadratic formula
We will use the quadratic formula to find the values of \(x\):

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

Substituting the values of \(a\), \(b\), and \(c\):

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} \]

This simplifies to:

\[ x = \frac{1 \pm \sqrt{1 + 56}}{4} \]

### Step 5: Simplify the expression under the square root
Now we simplify the expression under the square root:

\[ x = \frac{1 \pm \sqrt{57}}{4} \]

### Step 6: Write the final answers
Thus, the solutions for \(x\) are:

\[ x = \frac{1 + \sqrt{57}}{4} \quad \text{and} \quad x = \frac{1 - \sqrt{57}}{4} \]

### Summary of the solution:
The final answers are:

\[ x = \frac{1 + \sqrt{57}}{4} \quad \text{and} \quad x = \frac{1 - \sqrt{57}}{4} \]

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