Question 149395
If {{{D<0}}}, then the quadratic will have two complex (ie non real) solutions. 


For example, let's find the discriminant for {{{y=x^2+2x+5}}}



From {{{x^2+2x+5}}} we can see that {{{a=1}}}, {{{b=2}}}, and {{{c=5}}}



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



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



{{{D=4-4(1)(5)}}} Square {{{2}}} to get {{{4}}}



{{{D=4-20}}} Multiply {{{4(1)(5)}}} to get {{{(4)(5)=20}}}



{{{D=-16}}} Subtract {{{20}}} from {{{4}}} to get {{{-16}}}



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



Now let's use the quadratic formula to find the solutions of {{{y=x^2+2x+5}}}



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(2) +- sqrt( (2)^2-4(1)(5) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=2}}}, and {{{c=5}}}



{{{x = (-2 +- sqrt( 4-4(1)(5) ))/(2(1))}}} Square {{{2}}} to get {{{4}}}. 



{{{x = (-2 +- sqrt( 4-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{x = (-2 +- sqrt( -16 ))/(2(1))}}} Subtract {{{20}}} from {{{4}}} to get {{{-16}}}



{{{x = (-2 +- sqrt( -16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-2 +- 4*i)/(2)}}} Take the square root of {{{-16}}} to get {{{4*i}}}. 



{{{x = (-2 + 4*i)/(2)}}} or {{{x = (-2 - 4*i)/(2)}}} Break up the expression. 



{{{x = (-2)/(2) + (4*i)/(2)}}} or {{{x =  (-2)/(2) - (4*i)/(2)}}} Break up the fraction for each case. 



{{{x = -1+2*i}}} or {{{x =  -1-2*i}}} Reduce. 



So our answers are {{{x = -1+2*i}}} or {{{x = -1-2*i}}} 



Since our answers are complex, this verifies our original claim.