Question 318380


From {{{s^2-5s-8}}} we can see that {{{a=1}}}, {{{b=-5}}}, and {{{c=-8}}}



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



{{{D=(-5)^2-4(1)(-8)}}} Plug in {{{a=1}}}, {{{b=-5}}}, and {{{c=-8}}}



{{{D=25-4(1)(-8)}}} Square {{{-5}}} to get {{{25}}}



{{{D=25--32}}} Multiply {{{4(1)(-8)}}} to get {{{(4)(-8)=-32}}}



{{{D=25+32}}} Rewrite {{{D=25--32}}} as {{{D=25+32}}}



{{{D=57}}} Add {{{25}}} to {{{32}}} to get {{{57}}}



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