Question

solve for x in the equation $x^{2}+4x - 4 = 8$.\
\\(\bigcirc\\) $x = -6$ or $x = 2$\
\\(\bigcirc\\) $x = -2\pm2\sqrt{2}$\
\\(\bigcirc\\) $x = -2$ or $x = 6$\
\\(\bigcirc\\) $x = 2\pm2\sqrt{2}$

Explanation:

Step1: Rearrange equation to standard form

Subtract 8 from both sides to set the equation to 0:
$x^2 + 4x - 4 - 8 = 0$
$x^2 + 4x - 12 = 0$

Step2: Factor the quadratic expression

Find two numbers that multiply to -12 and add to 4: 6 and -2.
$(x + 6)(x - 2) = 0$

Step3: Solve for x using zero product rule

Set each factor equal to 0 and solve:
$x + 6 = 0 \implies x = -6$
$x - 2 = 0 \implies x = 2$

Answer:

$x = -6$ or $x = 2$