Question

16.) $2x^2 + 6x - 4 = 2$
18.) $x^2 + 8x - 12 = 3$

Explanation:

Step1: Solve equation 16: \(2x^2 + 6x - 4 = 2\)

First, rewrite the equation in standard quadratic form \(ax^2 + bx + c = 0\). Subtract 2 from both sides:
\(2x^2 + 6x - 6 = 0\). Divide all terms by 2: \(x^2 + 3x - 3 = 0\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), \(c = -3\).
Calculate the discriminant: \(b^2 - 4ac = 3^2 - 4(1)(-3) = 9 + 12 = 21\).
Substitute into the formula: \(x = \frac{-3 \pm \sqrt{21}}{2}\).

Step2: Solve equation 18: \(x^2 + 8x - 12 = 3\)

Rewrite in standard form: \(x^2 + 8x - 15 = 0\).

Use the quadratic formula with \(a = 1\), \(b = 8\), \(c = -15\).
Discriminant: \(b^2 - 4ac = 8^2 - 4(1)(-15) = 64 + 60 = 124\).
Simplify \(\sqrt{124} = 2\sqrt{31}\), so \(x = \frac{-8 \pm 2\sqrt{31}}{2} = -4 \pm \sqrt{31}\).

Answer:

• For \(2x^2 + 6x - 4 = 2\): \(x = \frac{-3 + \sqrt{21}}{2}\) or \(x = \frac{-3 - \sqrt{21}}{2}\)
• For \(x^2 + 8x - 12 = 3\): \(x = -4 + \sqrt{31}\) or \(x = -4 - \sqrt{31}\)