Question 154374
I don't know what to make of your work, but here's the way that I would do it:



{{{3^(1/2)y^2-4y-(3)^(1/2)=0 }}} Start with the given equation.



{{{sqrt(3)y^2-4y-sqrt(3)=0 }}} Rewrite {{{3^(1/2)}}} as {{{sqrt(3)}}}




Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=sqrt(3)}}}, {{{b=-4}}}, and {{{c=-sqrt(3)}}}




Let's use the quadratic formula to solve for y



{{{y = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{y = (-(-4) +- sqrt( (4)^2-4(sqrt(3))(-sqrt(3)) ))/(2*sqrt(3))}}} Plug in  {{{a=sqrt(3)}}}, {{{b=-4}}}, and {{{c=-sqrt(3)}}}



{{{y = (4 +- sqrt( (4)^2-4(sqrt(3))(-sqrt(3)) ))/(2*sqrt(3))}}} Negate -4 to get 4



{{{y = (4 +- sqrt( 16-4(sqrt(3))(-sqrt(3)) ))/(2*sqrt(3))}}} Square 4 to get 16



{{{y = (4 +- sqrt( 16-4(-3) ))/(2*sqrt(3))}}} Multiply {{{sqrt(3)}}} and {{{-sqrt(3)}}} to get {{{-sqrt(3)sqrt(3)=-sqrt(3*3)=-sqrt(3^2)=-3}}}




{{{y = (4 +- sqrt( 16+12 ))/(2*sqrt(3))}}} Multiply -4 and -3 to get 12



{{{y = (4 +- sqrt( 28 ))/(2*sqrt(3))}}} Add



{{{y = (4 +- 2*sqrt( 7 ))/(2*sqrt(3))}}} Simplify the square root



{{{y = (4 + 2*sqrt( 7 ))/(2*sqrt(3))}}}  or {{{y = (-4 - 2*sqrt( 7 ))/(2*sqrt(3))}}} Break up the "plus/minus"




{{{y = ((4 + 2*sqrt( 7 ))/(2*sqrt(3)))(sqrt(3)/sqrt(3))}}}  or {{{y = ((4 - 2*sqrt( 7 ))/(2*sqrt(3)))(sqrt(3)/sqrt(3))}}} Multiply both numerator and denominator by {{{sqrt(3)}}} to rationalize the denominator



{{{y = ((sqrt(3))(4 + 2*sqrt( 7 )))/(2*sqrt(3)sqrt(3))}}}  or {{{y = ((sqrt(3))(4 - 2*sqrt( 7 )))/(2*sqrt(3)sqrt(3))}}} Combine the fractions



{{{y = ((sqrt(3))(4 + 2*sqrt( 7 )))/(2*(3))}}}  or {{{y = ((sqrt(3))(4 - 2*sqrt( 7 )))/(2*(3))}}} Multiply



{{{y = ((sqrt(3))(4 + 2*sqrt( 7 )))/(6)}}}  or {{{y = ((sqrt(3))(4 - 2*sqrt( 7 )))/(6)}}} Multiply again



{{{y = (4*sqrt(3) + 2*sqrt(3)sqrt( 7 ))/(6)}}}  or {{{y = (4*sqrt(3) - 2*sqrt(3)sqrt( 7 ))/(6)}}} Distribute




{{{y = (4*sqrt(3) + 2*sqrt( 21 ))/(6)}}}  or {{{y = (4*sqrt(3) - 2*sqrt( 21 ))/(6)}}} Multiply 



{{{y = (2*sqrt(3) + sqrt( 21 ))/(3)}}}  or {{{y = (2*sqrt(3) - sqrt( 21 ))/(3)}}} Reduce




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Answer:



So the solutions are {{{y = (2*sqrt(3) + sqrt( 21 ))/(3)}}}  or {{{y = (2*sqrt(3) - sqrt( 21 ))/(3)}}}



which approximate to {{{y=2.68223}}} or {{{y=-0.37282}}}