Question

1. ( x^2 - 2x = 3 )
2. ( x^2 + 12x = -8 )

Explanation:

Step1: Solve \( x^2 - 2x = 3 \)

First, rewrite the equation in standard quadratic form \( ax^2 + bx + c = 0 \). Subtract 3 from both sides:
\( x^2 - 2x - 3 = 0 \)
Factor the quadratic. We need two numbers that multiply to -3 and add to -2. Those numbers are -3 and 1:
\( (x - 3)(x + 1) = 0 \)
Set each factor equal to zero and solve for \( x \):

• \( x - 3 = 0 \) gives \( x = 3 \)
• \( x + 1 = 0 \) gives \( x = -1 \)

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

Rewrite in standard form by adding 8 to both sides:
\( x^2 + 12x + 8 = 0 \)
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 12 \), \( c = 8 \).
First, calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = 12^2 - 4(1)(8) = 144 - 32 = 112 \)
Simplify \( \sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7} \)
Now apply the quadratic formula:
\( x = \frac{-12 \pm 4\sqrt{7}}{2(1)} = \frac{-12 \pm 4\sqrt{7}}{2} = -6 \pm 2\sqrt{7} \)

Answer:

1. For \( x^2 - 2x = 3 \), solutions are \( x = 3 \) and \( x = -1 \).
2. For \( x^2 + 12x = -8 \), solutions are \( x = -6 + 2\sqrt{7} \) and \( x = -6 - 2\sqrt{7} \).