How do you solve $( x + 5)² = 49$ ?

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Answer 1 · 2016-06-01T12:20:15

$x=-12$ or $x=2$

$(x+5)^2=49$

or $(x+5)^2-49=0$

or $(x+5)^2-7^2=0$

Now using $a^2-b^2=(a+b_(a-b)$

The above is equivalent to $((x+5)+7)((x+5)-7)=0$

or $(x+12)(x-2)=0$

Hence either $x+12=0$ or $x-2=0$

i.e. $x=-12$ or $x=2$

How do you solve ( x + 5)² = 49 ( x + 5)² = 49?