Question 360730
From {{{x^2+3x-4}}} we can see that {{{a=1}}}, {{{b=3}}}, and {{{c=-4}}}



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



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



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



{{{D=9--16}}} Multiply {{{4(1)(-4)}}} to get {{{(4)(-4)=-16}}}



{{{D=9+16}}} Rewrite {{{D=9--16}}} as {{{D=9+16}}}



{{{D=25}}} Add {{{9}}} to {{{16}}} to get {{{25}}}



So the discriminant is {{{D=25}}}



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



Note: because the discriminant is a perfect square, this means that the two real solutions are also two rational solutions.



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