Question 164994
I'll do the first two to get you started



Remember,

If D>0 (ie the discriminant is positive), then you will have 2 real solutions,
If D=0, then you will have only 1 real solution, or
If D<0 (ie the discriminant is negative), then you will have 2 complex solutions (ie you will have NO real solutions). 




a)



From {{{x^2+6x-7}}} we can see that {{{a=1}}}, {{{b=6}}}, and {{{c=-7}}}



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



{{{D=(6)^2-4(1)(-7)}}} Plug in {{{a=1}}}, {{{b=6}}}, and {{{c=-7}}}



{{{D=36-4(1)(-7)}}} Square {{{6}}} to get {{{36}}}



{{{D=36--28}}} Multiply {{{4(1)(-7)}}} to get {{{(4)(-7)=-28}}}



{{{D=36+28}}} Rewrite {{{D=36--28}}} as {{{D=36+28}}}



{{{D=64}}} Add {{{36}}} to {{{28}}} to get {{{64}}}



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



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b)



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



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



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



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



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



{{{D=-3}}} Subtract {{{4}}} from {{{1}}} to get {{{-3}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.