Please Help! 2x^2 - 7x + 12 = 0 ...?

2x^2-7x+12=0
Before beginning the "complete the square" process, the coefficient of the x^2 term MUST be +1
x^2-(7/2)x+6=0
x^2-(7/2)x= -6
x^2-(7/2)x+ ?= -6+?
? is always half the coefficient of the x term, squared. Add to both sides!
x^2-(7/2)x+49/16= -6+49/16
(x-7/4)^2= -96/16+49/16
(x-7/4)^2= -47/16
Now take the square root of both sides
x-7/4=+/-rt(-47) / 4
The rt(-47) poses a problem, because you cannot
take the square root of a negative number. We get
around that as follows
x-7/4=[rt(47)rt(-1)] / 4
Now rt(-1) = i
x-7/4= +/-rt(47)i / 4
x=7/4+/-rt(47)i/4
This is a Complex Number, and Complex Numbers
are written in the form a+bi, where a and b are Real
Numbers, and i =sq.rt.(-1)
Therefore we write our answer as
x=7/4+rt47)i/4, or
x=7/4-rt(47)i/4

Note. I knew I was headed for Complex roots, because the Discriminant of 2x^2-7x+12 is
b^2-4ac, where a=2, b=-7, c=12
b^2-4ac=(-7)^2-4(2)12
=49-96
=-47
It's negative, therefore Complex Solution only

x2 + 6x - 7 = (x + 7)(x - a million) x2 - 4x - 32 = (x - 8)(x + 4) 2x2 + 5x - 3 = (2x - a million)(x + 3) x2 + 5x - 14 = (x + 7)(x - 2) x2 + 3x -a million use the discriminant b2 - 4ac and replace the fee of a, b and c b2 - 4ac = 13, hence it has 2 authentic roots and irrational because of the fact 13 is greater beneficial than 0 yet not a eye-catching sq..

2 ( x^2 - (7/2) x ) = - 12

2 [ x^2 - (7/2)x + 49/16) ] = - 12 + 49/8

2 [ x - (7/4) ]^2 = - 47/8

[ x - (7/4) ]^2 = - 47 / 16

x - (7/4) = ± i √47/ 4

x = 7/4 ± i √47/ 4

x = (1/4) ( 7 ± i √47 )

2x^2 - 7x + 12 = 0
2x^2 - 7x = -12
(2x^2 - 7x)/2 = -12/2
x^2 - 7x/2 = -12/2
x^2 - (7x/2)/2 - (7x/2)/2 = -6
x^2 - (7x/2)(1/2) - (7x/2)(1/2) = -6
x^2 - 7x/4 - 7x/4 = -6
x^2 - 7x/4 - 7x/4 + 49/16 = -6 + 49/16
(x^2 - 7x/4) - (7x/4 - 49/16) = -96/16 + 49/16
x(x - 7/4) - 7/4(x - 7/4) = -47/16
(x - 7/4)(x - 7/4) = -47/16
(x - 7/4)^2 = -47/16
x - 7/4 = ±√(-47/16)
x = 7/4 ±i(√47)/4
x = (7 ±i√47)/4

Hi,

2x² - 7x + 12 = 0
2 [ x² - (7/2)x + 6 ] = 0
x² - (7/2)x + 6 = 0
x² - (7/2)x + [ (49/16) - (49/16) ] + 6 = 0
[ x² - (7/2)x + (49/16) ] - (49/16) + 6 = 0
[x - (7/4)]² - (49/16) + (96/16) = 0
[x - (7/4)]² + (47/16) = 0

Hope this helps!
J

x^2-(7/2)x+6=0
x^2-(7/2)+49/4=73/4
(x-7/2)^2=73/4
x-7/2=sqrt(73)/2 or x-7/2=-sqrt(73)/2
x=(7+sqrt(73))/2 or -(7+sqrt(73))/2

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