Question 259866
I'll do the first two to get you going


# 1



From {{{x^2+5x+6}}} we can see that {{{a=1}}}, {{{b=5}}}, and {{{c=6}}}



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



{{{D=(5)^2-4(1)(6)}}} Plug in {{{a=1}}}, {{{b=5}}}, and {{{c=6}}}



{{{D=25-4(1)(6)}}} Square {{{5}}} to get {{{25}}}



{{{D=25-24}}} Multiply {{{4(1)(6)}}} to get {{{(4)(6)=24}}}



{{{D=1}}} Subtract {{{24}}} from {{{25}}} to get {{{1}}}



Since the discriminant is greater than zero, this means that there are two real solutions.


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# 2



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



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



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



{{{D=16-4(3)(3)}}} Square {{{-4}}} to get {{{16}}}



{{{D=16-36}}} Multiply {{{4(3)(3)}}} to get {{{(12)(3)=36}}}



{{{D=-20}}} Subtract {{{36}}} from {{{16}}} to get {{{-20}}}



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



In other words, there are no real solutions.