Question 541323
From {{{3x^2+x+2}}} we can see that {{{a=3}}}, {{{b=1}}}, and {{{c=2}}}



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



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



{{{D=1-4(3)(2)}}} Square {{{1}}} to get {{{1}}}



{{{D=1-24}}} Multiply {{{4(3)(2)}}} to get {{{(12)(2)=24}}}



{{{D=-23}}} Subtract {{{24}}} from {{{1}}} to get {{{-23}}}



So the discriminant is {{{D=-23}}}



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



In other words, there are no real solutions.



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