Question 350043


{{{10y^2+y-2=0}}} Start with the given equation.



Notice that the quadratic {{{10y^2+y-2}}} is in the form of {{{Ay^2+By+C}}} where {{{A=10}}}, {{{B=1}}}, and {{{C=-2}}}



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



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



{{{y = (-(1) +- sqrt( (1)^2-4(10)(-2) ))/(2(10))}}} Plug in  {{{A=10}}}, {{{B=1}}}, and {{{C=-2}}}



{{{y = (-1 +- sqrt( 1-4(10)(-2) ))/(2(10))}}} Square {{{1}}} to get {{{1}}}. 



{{{y = (-1 +- sqrt( 1--80 ))/(2(10))}}} Multiply {{{4(10)(-2)}}} to get {{{-80}}}



{{{y = (-1 +- sqrt( 1+80 ))/(2(10))}}} Rewrite {{{sqrt(1--80)}}} as {{{sqrt(1+80)}}}



{{{y = (-1 +- sqrt( 81 ))/(2(10))}}} Add {{{1}}} to {{{80}}} to get {{{81}}}



{{{y = (-1 +- sqrt( 81 ))/(20)}}} Multiply {{{2}}} and {{{10}}} to get {{{20}}}. 



{{{y = (-1 +- 9)/(20)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{y = (-1 + 9)/(20)}}} or {{{y = (-1 - 9)/(20)}}} Break up the expression. 



{{{y = (8)/(20)}}} or {{{y =  (-10)/(20)}}} Combine like terms. 



{{{y = 2/5}}} or {{{y = -1/2}}} Simplify. 



So the solutions are {{{y = 2/5}}} or {{{y = -1/2}}} 

  

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