Question 193949


From {{{x^2+3x-15}}} we can see that {{{a=1}}}, {{{b=3}}}, and {{{c=-15}}}



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



{{{D=(3)^2-4(1)(-15)}}} Plug in {{{a=1}}}, {{{b=3}}}, and {{{c=-15}}}



{{{D=9-4(1)(-15)}}} Square {{{3}}} to get {{{9}}}



{{{D=9--60}}} Multiply {{{4(1)(-15)}}} to get {{{(4)(-15)=-60}}}



{{{D=9+60}}} Rewrite {{{D=9--60}}} as {{{D=9+60}}}



{{{D=69}}} Add {{{9}}} to {{{60}}} to get {{{69}}}



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