Question 419345


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



Notice we have a quadratic equation in the form of {{{an^2+bn+c}}} where {{{a=5}}}, {{{b=-7}}}, and {{{c=2}}}



Let's use the quadratic formula to solve for n



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



{{{n = (-(-7) +- sqrt( (-7)^2-4(5)(2) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=-7}}}, and {{{c=2}}}



{{{n = (7 +- sqrt( (-7)^2-4(5)(2) ))/(2(5))}}} Negate {{{-7}}} to get {{{7}}}. 



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



{{{n = (7 +- sqrt( 49-40 ))/(2(5))}}} Multiply {{{4(5)(2)}}} to get {{{40}}}



{{{n = (7 +- sqrt( 9 ))/(2(5))}}} Subtract {{{40}}} from {{{49}}} to get {{{9}}}



{{{n = (7 +- sqrt( 9 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{n = (7 +- 3)/(10)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{n = (7 + 3)/(10)}}} or {{{n = (7 - 3)/(10)}}} Break up the expression. 



{{{n = (10)/(10)}}} or {{{n =  (4)/(10)}}} Combine like terms. 



{{{n = 1}}} or {{{n = 2/5}}} Simplify. 



So the answers are {{{n = 1}}} or {{{n = 2/5}}} 

  

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