What are the x-intercepts of #y=x^2+2x-15#?

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Answer 1 · 2014-10-26T21:50:48

By setting it equal to zero,

#x^2+2x-15=0#

by factoring out,

#(x+5)(x-3)=0#

by setting each factor equal to zero,

#{(x+5=0 Rightarrow x=-5),(x-3=0 Rightarrow x=3):}#

Hence, the #x#-intercepts are: #x=-5,3#.

I hope that this was helpful.