Question 199098
I'll do the first three to get you started (they all follow the same logic)


a) 



From {{{x^2+3x-15}}} we can see that {{{a=1}}}, {{{b=3}}}, and {{{c=-15}}}



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



{{{D=(3)^2-4(1)(-15)}}} Plug in {{{a=1}}}, {{{b=3}}}, and {{{c=-15}}}



{{{D=9-4(1)(-15)}}} Square {{{3}}} to get {{{9}}}



{{{D=9--60}}} Multiply {{{4(1)(-15)}}} to get {{{(4)(-15)=-60}}}



{{{D=9+60}}} Rewrite {{{D=9--60}}} as {{{D=9+60}}}



{{{D=69}}} Add {{{9}}} to {{{60}}} to get {{{69}}}



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


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b)




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



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



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



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



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



{{{D=-15}}} Subtract {{{16}}} from {{{1}}} to get {{{-15}}}



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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c)





From {{{x^2-4x-7}}} we can see that {{{a=1}}}, {{{b=-4}}}, and {{{c=-7}}}



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



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



{{{D=16-4(1)(-7)}}} Square {{{-4}}} to get {{{16}}}



{{{D=16--28}}} Multiply {{{4(1)(-7)}}} to get {{{(4)(-7)=-28}}}



{{{D=16+28}}} Rewrite {{{D=16--28}}} as {{{D=16+28}}}



{{{D=44}}} Add {{{16}}} to {{{28}}} to get {{{44}}}



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