Question 154211


{{{(y+2)^2=20}}} Start with the given equation.



{{{x+2=0+-sqrt(20)}}} Take the square root of both sides.



{{{y+2=sqrt(20)}}} or {{{y+2=-sqrt(20)}}} Break up the "plus/minus" to form two equations.



{{{y+2=2*sqrt(5)}}} or {{{y+2=-2*sqrt(5)}}}  Simplify the square root.



{{{y=-2+2*sqrt(5)}}} or {{{y=-2-2*sqrt(5)}}} Subtract {{{2}}} from both sides.



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Answer:



So the solutions are {{{y=-2+2*sqrt(5)}}} or {{{y=-2-2*sqrt(5)}}}.



which approximate to {{{y=2.472}}} or {{{y=-6.472}}}