Question

solve for (x) in the equation (2x^2 + 3x - 7 = x^2 + 5x + 39).

\\(x = -6 \pm \sqrt{82}\\)
\\(x = -6 \pm 2\sqrt{17}\\)
\\(x = 1 \pm \sqrt{33}\\)
\\(x = 1 \pm \sqrt{47}\\)

Explanation:

⚡ Using what you learned: factoring and solving quadratic equations

Step 1: Move all terms to one side

$$ 2x^2 + 3x - 7 = x^2 + 5x + 39 $$

Subtract \(x^2\), \(5x\), and \(39\) from both sides:

$$ x^2 - 2x - 46 = 0 $$

Step 2: Apply the quadratic formula

For \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -2\), and \(c = -46\):

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

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

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

$$ x = \frac{2 \pm \sqrt{188}}{2} $$

Step 3: Simplify the radical

Simplify \(\sqrt{188}\):

$$ \sqrt{188} = \sqrt{4 \times 47} = 2\sqrt{47} $$

Substitute back into the expression:

$$ x = \frac{2 \pm 2\sqrt{47}}{2} $$

$$ x = 1 \pm \sqrt{47} $$

Answer:

\(x = 1 \pm \sqrt{47}\)