Answer:
The solutions are [tex]x=2,\:x=-6[/tex]
Step-by-step explanation:
To solve for [tex]x[/tex] the equation [tex]x^2+4x-4=8[/tex] you must:
[tex]\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides}\\\\x^2+4x-4-8=8-8\\\\x^2+4x-12=0[/tex]
[tex]\mathrm{Factor\:}x^2+4x-12\\\\\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(x^2-2x\right)+\left(6x-12\right)\\\\x\left(x-2\right)+6\left(x-2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x-2\\\\x\left(x-2\right)+6\left(x-2\right)[/tex]
Using the Principle of Zero Products.
If ab = 0, then either a = 0 or b = 0, or both a and b are 0.
[tex]x-2=0\\x=2[/tex]
[tex]x+6=0\\x=-6[/tex]
The solutions are [tex]x=2,\:x=-6[/tex].