Question 176363


{{{y^2-4y+4=36}}} Start with the given equation.



{{{y^2-4y+4-36=0}}} Subtract 36 from both sides.



{{{y^2-4y-32=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=1}}}, {{{b=-4}}}, and {{{c=-32}}}



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(1)(-32) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-4}}}, and {{{c=-32}}}



{{{y = (4 +- sqrt( (-4)^2-4(1)(-32) ))/(2(1))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{y = (4 +- sqrt( 16-4(1)(-32) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{y = (4 +- sqrt( 16--128 ))/(2(1))}}} Multiply {{{4(1)(-32)}}} to get {{{-128}}}



{{{y = (4 +- sqrt( 16+128 ))/(2(1))}}} Rewrite {{{sqrt(16--128)}}} as {{{sqrt(16+128)}}}



{{{y = (4 +- sqrt( 144 ))/(2(1))}}} Add {{{16}}} to {{{128}}} to get {{{144}}}



{{{y = (4 +- sqrt( 144 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (4 +- 12)/(2)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{y = (4 + 12)/(2)}}} or {{{y = (4 - 12)/(2)}}} Break up the expression. 



{{{y = (16)/(2)}}} or {{{y =  (-8)/(2)}}} Combine like terms. 



{{{y = 8}}} or {{{y = -4}}} Simplify. 



So the answers are {{{y = 8}}} or {{{y = -4}}} 

  

Note: the exact solutions and the approximate solutions are the same (since no approximation is not needed).