Question 512143
From {{{19x^2-5x+4}}} we can see that {{{a=19}}}, {{{b=-5}}}, and {{{c=4}}}



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



{{{D=(-5)^2-4(19)(4)}}} Plug in {{{a=19}}}, {{{b=-5}}}, and {{{c=4}}}



{{{D=25-4(19)(4)}}} Square {{{-5}}} to get {{{25}}}



{{{D=25-304}}} Multiply {{{4(19)(4)}}} to get {{{(76)(4)=304}}}



{{{D=-279}}} Subtract {{{304}}} from {{{25}}} to get {{{-279}}}



So the discriminant is {{{D=-279}}}



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



In other words, there are no real solutions.