document.write( "Question 389988: Find the solution set of x^2+2x-8>0\r
\n" ); document.write( "\n" ); document.write( "a. {x: -4\n" ); document.write( "b. {x: x<-4 or x>2}
\n" ); document.write( "c. {x: -2\n" ); document.write( "d. {x: x<-2 or x>4}\r
\n" ); document.write( "\n" ); document.write( "I know that answer is b.
\n" ); document.write( "But I don't get how to get that.
\n" ); document.write( "Please solve it.
\n" ); document.write( "

Algebra.Com's Answer #276461 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Hi, \r
\n" ); document.write( "\n" ); document.write( "Note: x^2+2x-8 = 0 when x = 2, x = -4. Note: Factors are (x+4)(x-2)
\n" ); document.write( "for x^2+2x-8 > 0 , then {x: x<-4 or x>2}
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