Answer:
[tex]x=12,\:x=-6[/tex] are your x intercepts
Step-by-step explanation:
[tex]\left(x-12\right)\left(x+6\right)=0\\\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)\\\mathrm{Solve\:}\:x-12=0:\quad x=12\\x-12=0\\\mathrm{Add\:}12\mathrm{\:to\:both\:sides}\\x-12+12=0+12\\Simplify\\x=12\\\mathrm{Solve\:}\:x+6=0:\quad x=-6\\x+6=0\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\x+6-6=0-6\\Simplify\\x=-6\\The\:solutions\:to\:the\:quadratic\:equation\:are:\\x=12,\:x=-6[/tex]
