Question 197006


From {{{x^2-3x+4}}} we can see that {{{a=1}}}, {{{b=-3}}}, and {{{c=4}}}



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



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



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



{{{D=9-16}}} Multiply {{{4(1)(4)}}} to get {{{(4)(4)=16}}}



{{{D=-7}}} Subtract {{{16}}} from {{{9}}} to get {{{-7}}}



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



In other words, there are no <i>real</i> solutions.