Question 378093


{{{2y^2-50=0}}} Start with the given equation.



{{{2y^2=0+50}}}Add {{{50}}} to both sides.



{{{2y^2=50}}} Combine like terms.



{{{y^2=(50)/(2)}}} Divide both sides by {{{2}}}.



{{{y^2=25}}} Reduce.



{{{y=""+-sqrt(25)}}} Take the square root of both sides.



{{{y=sqrt(25)}}} or {{{y=-sqrt(25)}}} Break up the "plus/minus" to form two equations.



{{{y=5}}} or {{{y=-5}}}  Take the square root of {{{25}}} to get {{{5}}}.



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Answer:



So the solutions are {{{y=5}}} or {{{y=-5}}}.




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