Question 172399
Note: {{{(3)^(1/2)=sqrt(3)}}}



So {{{(3)^(1/2)y^2 - 4y - 7(3)^(1/2)y = 0}}} also looks like {{{sqrt(3)y^2 - 4y - 7sqrt(3)y = 0}}}




From {{{sqrt(3)y^2 - 4y - 7sqrt(3)y}}} we can see that {{{a=sqrt(3)}}}, {{{b=-4}}}, and {{{c=7sqrt(3)}}}



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



{{{D=(-4)^2-4(sqrt(3))(-7sqrt(3))}}} Plug in {{{a=sqrt(3)}}}, {{{b=-4}}}, and {{{c=7sqrt(3)}}}



{{{D=16-4(sqrt(3))(-7sqrt(3))}}} Square {{{-4}}} to get {{{16}}}



{{{D=16+84}}} Multiply {{{-4(sqrt(3))(-7sqrt(3))}}} to get {{{(-4)*(-7)*sqrt(3*3)=28*sqrt(9)=28*3=84}}}



{{{D=100}}} Add




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