How do you solve $2( 3- y ) ^ { 2} = 18$ ?

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Answer 1 · 2017-05-14T15:42:03

${0;6}$

$2*(3−y)^2=18 <=> (3-y)^2 = 18/2$

$<=> (3-y)^2 = 9$

$<=> 3-y = 3$ or $3-y = -3$

$<=> y=0$ or $y = 6$

How do you solve 2( 3- y ) ^ { 2} = 182( 3- y ) ^ { 2} = 18?