Question 145005


From {{{x^2+3x+10}}} we can see that {{{a=1}}}, {{{b=3}}}, and {{{c=10}}}



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



{{{D=(3)^2-4(1)(10)}}} Plug in {{{a=1}}}, {{{b=3}}}, and {{{c=10}}}



{{{D=9-4(1)(10)}}} Square {{{3}}} to get {{{9}}}



{{{D=9-40}}} Multiply {{{4(1)(10)}}} to get {{{(4)(10)=40}}}



{{{D=-31}}} Subtract {{{40}}} from {{{9}}} to get {{{-31}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.