solve by completing the square?

x² + 2x = 3 →
x² + 2x - 3 = 0
rewrite -3 as 1 - 4:
x² + 2x +1 - 4 = 0 →
(x² + 2x +1) - 4 = 0 →
(x +1)² - 4 = 0 →
now factor the left side as a difference between two squares:
[(x +1) - 2][(x +1) + 2] = 0 →
(x +1 - 2)(x +1 + 2) = 0 →
(x - 1)(x + 3) = 0 →
thus your solutions are: (x - 1) = 0 (x + 3) = 0
(x - 1) = 0 → x = 1
(x + 3) = 0 → x = -3

x² + 8x = 7 →
x² + 8x - 7 = 0
8x is the double product of the square, thus, being the known term x of the square, the second term has to be (8x/x)(1/2) = 4;
then, being 4² = 16, you have to rewrite -7 as 16 - (7 +16) = 16 - 23:
x² + 8x +16 - 23 = 0 →
(x² + 8x +16) - 23 = 0 →
(x + 4)² - 23 = 0 →
let us factor left side as a difference between squares:
[(x + 4) - √23][(x + 4) +√23] = 0 →
[(x + (4 - √23)][x + (4 +√23)] = 0 →
thus the solutions are:
[(x + (4 - √23)] = 0 → x = - (4 - √23) = - 4 + √23
[(x + (4 +√23)] = 0 → x = - (4 + √23) = - 4 - √23

x² + 12x = 11 →
x² + 12x - 11 = 0 →
being 12x the double product of the required square, and being x the known term of the square, the remaining term has to be:
(12x/x)(1/2) = 6
therefore, being 6²= 36, you have to rewrite -11 as
36 - 11 - 36 = 36 - (11 + 36) = 36 - 47:
x² + 12x + 36 - 47= 0 →
(x² + 12x + 36) - 47= 0 →
(x + 6)² - 47= 0 →
then factor the left side as a difference between squares:
[(x + 6) - √47][(x + 6) + √47] = 0 →
(x + 6 - √47)(x + 6 + √47) = 0 →
[(x + (6 - √47)][x + (6 + √47)] = 0 →
therefore the solutions are:
[(x + (6 - √47)] = 0 → x = - (6 - √47) = - 6 + √47
[x + (6 + √47)] = 0 → x = - (6 +√47) = - 6 - √47

I hope it has been helpful
Bye!

question 1
x² + 2x + 1 = 4
(x + 1)² = 4
(x + 1) = ± 2
x = - 1 ± 2
x = 1 , x = - 3

Question 2
x² + 8x + 16 = 23
(x + 4)² = 23
x + 4 = ± √23
x = - 4 ± √ 23

Question 3
x² + 12x + 36 = 47
(x + 6)² = 47

x + 6 = ± √ 47
x = - 6 ± √ 47

1) x^2 + 2x=3
dividing the equation by coefficient of x^2, we got
x^2+2x=3
adding both sides with (1/2 of coefficient of x)^2, we got
x^2+2x+(1/2of2)^2= 3+(1/2of2)^2
x^2+2x+1 = 3+1
(x+1)^2=4
taking sq roots of both sides
x+1=+ - 2
x= -2-1 andx=2-1
x= -3and 1

2)x^2 + 8x = 7
dividing the equation by coefficient of x^2, we got
x^2 + 8x = 7
adding both sides with (1/2 of coefficient of x)^2, we got
x^2+8x+(1/2of8)^2 = 7+(1/2of8)^2
x^2+8x+16 = 7+16
(x + 4)^2 = 23
x + 4 = ±√23
x + 4 = ±4.79
x = 4.79 - 4 and x = - 4.79 - 4
x = 0.79 and - 8.79

3) x^2 + 12x = 11
dividing the equation by coefficient of x^2, we got
x^2 + 12x = 11
adding both sides with (1/2 of coefficient of x)^2, we got
x^2 +12x + (1/2of12)^2 = 11 + (1/2of12)^2
x^2 + 12x + 36=11 + 36
(x + 6)^2 = 47
x + 6 = ±√47
x + 6 = ±6.85
x + 6 = 6.85 and x + 6 = - 6.85
x = 6.85 - 6 and x = - 6.85 - 6
x = 0.85 and - 12.85

add -3 to both sides of the equation
x^2 + 2x - 3 = 0
(x+3)*(x-1) = 0 then x = -3 or 1
add 16 to both sides of the equation
x^2 + 8x + 16 = 23
(x+4)*(x+4) = 23 then x = ((23)^(-1/2)) - 4
add 36 to both sides of the equation
x^2 + 12x + 36 = 47
(x+6)*(x+6) = 47 then x = ((47)^(-1/2)) - 6
You can also use the quadratic equation formula for
x^2 + 8x - 7 = 0
and
x^2 + 12x - 11 = 0

x^2+2x=3
x^2+(2x/2)+(2/2)=3+(2/2)
(x+1)^2=4
(x+1)=+- 2
x=+-2 -1
x=1, x=-3

x² + 2x = 3
To complete square x² + 2x
2nd term = 2 × √(1st term)(3rd term)
2 x = 2 x √(3rd term)
√(3rd term) = 1
(3rd term) = 1
x² + 2x + 1 = 3 + 1
(x + 1)² = 4 = 2²
(x + 1) = ± 2
x = – 1 + 2 = 1
or x = – 1 – 2 = – 3

x² + 8x = 7
x² + 8x + 16 = 7 + 16
(x + 4)² = 23
(x + 4) = √23
x = – 4 ± √23

x² + 12x = 11
x² + 12x + 36 = 11 + 36
(x + 6)² = 47
x + 6 = √47
x = – 6 ± √47
---------------

1) x^2 + 2x = 3

(x + 1)^2 - 2 = 3

(x + 1)^2 = 5

(x + 1) = SQRT(5)

x = -1 ± SQRT(5)

You can then bang this into a calculator to finnish it.

2) x^2 + 8x = 7

(x + 4)^2 - 16 = 7

(x + 4)^2 = 23

(x + 4) = SQRT(23)

x = -4 ± SQRT(23)

3) x^2 + 12x = 11

(x + 6)^2 - 36 = 11

(x + 6)^2 = 47

(x + 6) = SQRT(47)

x = -6 ± SQRT(47)

Hope this is helpful.

when you have x² + bx, you take half of b and square it to get the square of (x + b/2)², like this:

x² + 2x = 3
x² + 2x + 1 = 3 + 1
(x + 1)² = 4
x + 1 = 2 or x + 1 = -2
x = 1 or x = -3

the last one looks like:
x² + 12x = 11
x² + 12x + 6² = 11 + 6²
(x+6)² = 11 + 36
x + 6 = ±√47
x = -6 ±√47

x^2+2x=3
x^2+2x+1=3+1, take half of b (which is 2) and square it, then add that sum to both sides
(x+1)^2=4, the first part factors to this
Take the square root of both sides and get: x+1=+-2, where x=-3 and 1

x^2+8x=7
x^2+8x+16=7+16
(x+4)^2=23
x+4=+-sqrroot of 23
x=-4+-sqrroot of 23

x^2+12x=11
x^2+12x+36=47
(x+6)^2=47
x=-6+-sqrroot of 47

1)
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 1)^2 - 4 = 0
(x + 1)^2 = 4
x + 1 = ±√4
x + 1 = ±2

x + 1 = 2
x = 2 - 1
x = 1

x + 1 = -2
x = -2 - 1
x = -3

∴ x = 1 , -3

= = = = = = = =

2)
x^2 + 8x = 7
x^2 + 8x - 7 = 0
(x + 4)^2 - 23 = 0
(x + 4)^2 = 23
x + 4 = ±√23
x + 4 = ±4.79

x + 4 = 4.79
x = 4.79 - 44
x = 0.79

x + 4 = -4.79
x = -4.79 - 4
x = -8.79

∴ x = 0.79 , -8.79

= = = = = = = =

3)
x^2 + 12x = 11
x^2 + 12x - 11 = 0
(x + 6)^2 - 47 = 0
(x + 6)^2 = 47
x + 6 = ±√47
x + 6 = ±6.85

x + 6 = 6.85
x = 6.85 - 6
x = 0.85

x + 6 = -6.85
x = -6.85 - 6
x = -12.85

∴ x = 0.85 , 12.85