solve for x in the equation $x^{2}+4x - 4 = 8$.\n$x=-6$ or $x = 2$\n$x=-2pm2sqrt{2}$\n$x=-2$ or $x =…
solve for x in the equation $x^{2}+4x - 4 = 8$.\n$x=-6$ or $x = 2$\n$x=-2pm2sqrt{2}$\n$x=-2$ or $x = 6$\n$x=2pm2sqrt{2}$
Answer
Explanation:
Step1: Rearrange to standard quadratic form
First, rewrite the equation $x^{2}+4x - 4=8$ as $x^{2}+4x-12 = 0$.
Step2: Factor the quadratic equation
We factor $x^{2}+4x - 12$ into $(x + 6)(x - 2)=0$.
Step3: Solve for x
Set each factor equal to zero: If $x+6=0$, then $x=-6$; if $x - 2=0$, then $x = 2$.
Answer:
$x=-6$ or $x = 2$