Question 221485


From {{{x^2-18}}} we can see that {{{a=1}}}, {{{b=0}}}, and {{{c=-18}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=(0)^2-4(1)(-18)}}} Plug in {{{a=1}}}, {{{b=0}}}, and {{{c=-18}}}



{{{D=0-4(1)(-18)}}} Square {{{0}}} to get {{{0}}}



{{{D=0--72}}} Multiply {{{4(1)(-18)}}} to get {{{(4)(-18)=-72}}}



{{{D=0+72}}} Rewrite {{{D=0--72}}} as {{{D=0+72}}}



{{{D=72}}} Add {{{0}}} to {{{72}}} to get {{{72}}}



Since the discriminant is greater than zero, this means that there are two real solutions.