Question 353887
I'm assuming you meant to say {{{5y^2+7y-23=0}}}



First, let's solve {{{5y^2+7y-23=0}}}



{{{5y^2+7y-23=0}}} Start with the given equation.



Notice that the quadratic {{{5y^2+7y-23}}} is in the form of {{{Ay^2+By+C}}} where {{{A=5}}}, {{{B=7}}}, and {{{C=-23}}}



Let's use the quadratic formula to solve for "y":



{{{y = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{y = (-(7) +- sqrt( (7)^2-4(5)(-23) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=7}}}, and {{{C=-23}}}



{{{y = (-7 +- sqrt( 49-4(5)(-23) ))/(2(5))}}} Square {{{7}}} to get {{{49}}}. 



{{{y = (-7 +- sqrt( 49--460 ))/(2(5))}}} Multiply {{{4(5)(-23)}}} to get {{{-460}}}



{{{y = (-7 +- sqrt( 49+460 ))/(2(5))}}} Rewrite {{{sqrt(49--460)}}} as {{{sqrt(49+460)}}}



{{{y = (-7 +- sqrt( 509 ))/(2(5))}}} Add {{{49}}} to {{{460}}} to get {{{509}}}



{{{y = (-7 +- sqrt( 509 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{y = (-7+sqrt(509))/(10)}}} or {{{y = (-7-sqrt(509))/(10)}}} Break up the expression.  



So the solutions in terms of 'y' are {{{y = (-7+sqrt(509))/(10)}}} or {{{y = (-7-sqrt(509))/(10)}}} 




Now recall that you let {{{y=(3x-4)^3}}}. So this means that 



{{{(3x-4)^3 = (-7+sqrt(509))/(10)}}} or {{{(3x-4)^3 = (-7-sqrt(509))/(10)}}} 




Start with the given pair of equations


{{{(3x-4)^3 = (-7+sqrt(509))/(10)}}} or {{{(3x-4)^3 = (-7-sqrt(509))/(10)}}} 



Take the cube root of both sides to get: 


{{{3x-4 =root(3, (-7+sqrt(509))/(10))}}} or {{{3x-4 =root(3, (-7-sqrt(509))/(10))}}}


Note: I'm only taking the principal cube root 



Add 4 to both sides to get: 

{{{3x =root(3, (-7+sqrt(509))/(10))+4}}} or {{{3x=root(3, (-7-sqrt(509))/(10))+4}}}



Divide both sides by 3 to get:  

{{{x =(root(3, (-7+sqrt(509))/(10))+4)/3}}} or {{{x=(root(3, (-7-sqrt(509))/(10))+4)/3}}}



So the two real solutions are 



{{{x =(root(3, (-7+sqrt(509))/(10))+4)/3}}} or {{{x=(root(3, (-7-sqrt(509))/(10))+4)/3}}}