Question

Find (x+3)^2=16 , find x . Answer fast please ​

Answer 1

Answer:

(x+3)^2=16

x^2+6x+9-16=0

x^2+6x-7=0

x^2+7x-x-7=0

x(x+7)-1(x+7)=0

(x+7)(x-1)=0

x= -7,x=1

Answer 2

ANSWER,

$\sf\implies (x+3)^2=16$

$\sf\implies x^2+ 3^2+ 2 \times (3)(x)=16$

$\sf\implies x^2+9+6x=16$

$\sf\implies x^2+9+6x-16=0$

$\sf\implies x^2+6x-7=0$

$\sf\implies x^2-7x+x-7=0$

$\sf\implies x(x-7)+ 1(x-7)=0$

$\sf\implies  (x+!)(x-7)=0$

$\sf\implies  x+1=0,x-7=0$

$\sf\implies x=-1,x=7$

$\sf\implies  x=-1,7$

$\large\underline\bold\star x=-1,7$

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