Question 194753


From {{{t^2+8t+20}}} we can see that {{{a=1}}}, {{{b=8}}}, and {{{c=20}}}



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



{{{D=(8)^2-4(1)(20)}}} Plug in {{{a=1}}}, {{{b=8}}}, and {{{c=20}}}



{{{D=64-4(1)(20)}}} Square {{{8}}} to get {{{64}}}



{{{D=64-80}}} Multiply {{{4(1)(20)}}} to get {{{(4)(20)=80}}}



{{{D=-16}}} Subtract {{{80}}} from {{{64}}} to get {{{-16}}}



Since the discriminant is less than zero, this means that there are two complex solutions. 



In other words, there are no real solutions.