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

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Answer 1 · 2017-05-14T15:42:03

${0; 6}$ ${0;6}$

$2 \cdot {(3 - y)}^{2} = 18 \Leftrightarrow {(3 - y)}^{2} = \frac{18}{2}$ $2*(3−y)^2=18 <=> (3-y)^2 = 18/2$

$\Leftrightarrow {(3 - y)}^{2} = 9$ $<=> (3-y)^2 = 9$

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

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

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