Solve for x in the equation 2 x squared + 3 x minus 7 = x squared + 5 x + 39.
x = negative 6 plus-or-minus StartRoot 82 EndRoot
x = negative 6 plus-or-minus 2 StartRoot 17 EndRoot
x = 1 plus-or-minus StartRoot 33 EndRoot
x = 1 plus-or-minus StartRoot 47 EndRoot

We are given the equation:

\[
2x^2 + 3x - 7 = x^2 + 5x + 39
\]

### Step 1: Rearrange the equation
Move all terms to one side to set the equation to zero:

\[
2x^2 + 3x - 7 - (x^2 + 5x + 39) = 0
\]

Simplify the equation:

\[
2x^2 + 3x - 7 - x^2 - 5x - 39 = 0
\]

\[
(x^2 - 2x - 46) = 0
\]

### Step 2: Use the quadratic formula
We have a quadratic equation of the form:

\[
x^2 - 2x - 46 = 0
\]

The quadratic formula is:

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

For our equation, \(a = 1\), \(b = -2\), and \(c = -46\). Plug these values into the quadratic formula:

\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-46)}}{2(1)}
\]

Simplify:

\[
x = \frac{2 \pm \sqrt{4 + 184}}{2}
\]

\[
x = \frac{2 \pm \sqrt{188}}{2}
\]

Since \(\sqrt{188} = \sqrt{4 \times 47} = 2\sqrt{47}\), we have:

\[
x = \frac{2 \pm 2\sqrt{47}}{2}
\]

### Step 3: Final simplification
Simplify the expression:

\[
x = 1 \pm \sqrt{47}
\]

Thus, the solution is:

\[
x = 1 \pm \sqrt{47}
\]

This matches the option:

\[
x = 1 \pm \sqrt{47}
\]

### Would you like further explanation or details on this solution?

Here are 5 related questions:
1. What is the derivation of the quadratic formula?
2. How can we determine if a quadratic equation has real solutions?
3. What are the applications of the discriminant in quadratic equations?
4. Can you factor the quadratic \(x^2 - 2x - 46\) by completing the square?
5. What is the significance of the plus-or-minus symbol in the solution?

**Tip:** Always check your quadratic discriminant \((b^2 - 4ac)\) to determine if the solutions are real, complex, or repeated.

Solve for x in the equation 2x^2 + 3x - 7 = x^2 + 5x + 39.

Learn how to solve the quadratic equation 2x^2 + 3x - 7 = x^2 + 5x + 39 with a step-by-step explanation. This problem involves simplifying terms and applying the quadratic formula, perfect for students in Grades 9-10. The final solution is x = 1 ± √47.

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