find the solution set in a+bi form: x^2+6x+12=0?

x^2+6x+12=0 this is not factorable, so we will use completing the square

x^2+(6/2)^2x=-12
x^2+ 6x+9=-12+9
(x+3)^2=-3
x+3=+/-√[3*-1]
x=-3+/-i√[3]

x=-3+i√[3] or x=-3-i√[3] answer @(^_^)@

x² + 6x + 12 = 0
x² + 3x = - 12 + 3²
x² + 3x = - 12 + 9
(x + 3)² = - 3
x + 3 = 3i

Factors:
= x + 3 - 3i
= x + 3 + 3i

x = 3i - 3
x = - 3i - 3

Answer: x = 3i - 3, - 3i - 3; Factors: (x + 3 - 3i)(x - 3i - 3)

x = [ - 6 ± √ (36 - 48) ] / 2

x = [ - 6 ± √ ( - 12 ) ] / 2

x = [ - 6 ± i √ ( 12 ) ] / 2

x = [ - 6 ± 2 i √ 3 ] / 2

x = - 3 ± i √ 3

x^2+6x+12=0
D=36-4*1*12=-12=12*i^2
x1=(-6+2i√3)/2=-3+i√3
x2=(-6-2i√3)/2=-3-i√3

First complete the square :

x² + 6x + 12 = 0
x² + 6x = -12

Divide the second term by 2, then square :
6/2 = 3
3² = 9

x² + 6x + 9 = -12 + 9
(x + 3)² = -3
x + 3 = √-3
x + 3 = i√3
x = -3 ± i√3

x² + 6x + 12 = 0
x² + 6x = -12
x² + 6x + 9 = -12 + 9
(x + 3)² = -3
x + 3 = ±√3 i
x = -3 ± √3 i

x^2 + 6x + 12 = 0 Subtract 12 from both aspect : x^2 + 6x = -12 Calculate 0.5 of the 6 (= 3), then sq. it (3^2 = 9). Now upload this 9 to both aspect : x^2 + 6x + 9 = -12 + 9 = -3 The left-hand aspect is now a sq. : (x + 3)^2 = -3 Take the sq. root of both aspect : x + 3 = ± sqrt(-3) = ± sqrt(3)*i Subtract 3 from both aspect : x = -3 ± sqrt(3)*i, that is now contained in the style a ± bi. this technique is termed "polishing off the sq.".

Use either completing the square or quadratic formula

Question Number 1 :
For this equation x^2 + 6*x + 12 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !
B. Use completing the square to find the root of the equation !

Answer Number 1 :
The equation x^2 + 6*x + 12 = 0 is already in a*x^2+b*x+c=0 form.
As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = 6, c = 12.

1A. Find the roots using Quadratic Formula !
Use the formula,
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
As we know that a = 1, b = 6 and c = 12,
we need to subtitute a,b,c in the abc formula, with thos values.
So we get x1 = (-(6) + sqrt( (6)^2 - 4 * (1)*(12)))/(2*1) and x2 = (-(6) - sqrt( (6)^2 - 4 * (1)*(12)))/(2*1)
Which can be turned into x1 = ( -6 + sqrt( 36-48))/(2) and x2 = ( -6 - sqrt( 36-48))/(2)
Which make x1 = ( -6 + sqrt( -12))/(2) and x2 = ( -6 - sqrt( -12))/(2)
Which can be turned into x1 = ( -6 + sqrt(12)*sqrt(-1))/(2) and x2 = ( -6 - sqrt(12)*sqrt(-1))/(2)
As sqrt(-1) = i,
So we get x1 = ( -6 + 3.46410161513775*i )/(2) and x2 = ( -6 - 3.46410161513775*i )/(2)
So we have the answers x1 = -3 + 1.73205080756888*i and x2 = -3 - 1.73205080756888*i

1B. Use completing the square to find the root of the equation !
x^2 + 6*x + 12 = 0 ,divide both side with 1
So we get x^2 + 6*x + 12 = 0 ,
Which means that the coefficient of x is 6
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = 6/2 = 3
So we have make the equation into x^2 + 6*x + 9 + 3 = 0
Which can be turned into ( x + 3 )^2 + 3 = 0
Which can be turned into (( x + 3 ) - 1.73205080756888*i ) * (( x + 3 ) + 1.73205080756888*i ) = 0
By opening the brackets we will get ( x + 3 - 1.73205080756888*i ) * ( x + 3 + 1.73205080756888*i ) = 0
So we got the answers as x1 = -3 - 1.73205080756888*i and x2 = -3 + 1.73205080756888*i

x^2 + 6x + 12 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 6
c = 12

x = [-6 ±√(36 - 48)]/2
x = [-6 ±√(-12)]/2
x = [-6 ±√(2^2 * -1 * 3)]/2
x = [-6 ±2i√3]/2
x = -3 ±i√3

∴ x = -3 ±i√3