Verified answer

x= -b ± √b^2 - 4ac ALL OVER (2a)

x= 4 ± √16 - 4(2)(3)

ALL OVER 2(2)

x= 4 ± √-8

ALL OVER 4

x= 4 ± 2i√2

ALL OVER 4

x= 2 ± i√2

ALL OVER 2

They're actually both the same value. Using the quadratic formula gives

[ -b ±√(b^2 - 4ac) ] / 2a

[ 4 ±√(16 - 4*2*3) ] / (2*2)

[ 4 ±√(16 - 24) ] / 4

[ 4 ± √-8 ] / 4

[ 4 ± √(-2*4) ] / 4

[ 4 ± 2√-2 ] / 4

[ 4 ± 2i√2 ] / 4

If you decide to fully distribute the 4 in the denominator, you get 1 ± ((i√2) / 2), which is what you yourself got. If you divide the top and bottom by 2, you get what's in the book.

2x^2 - 4x + 3 = 0

a = 2

b = -4

c = 3

then it is the opposite of b plus and minus the square root of b^2 - 4ac all divided by 2a so plug and chug!

4 +- SR of -4^2 - 4(2)(3)/ 2(2)

4 = - SR of 16 - 24/4

4 plus and minus the square root of -8/4

No solution because you can not square root a negative number.

I am not sure how they came up with the answer in the back of the book.

Using the formula, we get x = (4 +/- sqrt(16-24))/4, or (2 +/- sqrt(-2))/2. Both you and the book are correct.

x = (-b+-sqrt(b^2-4ac))/(2a)

= (-(-4)+-sqrt(16-24))/(2*2)

= (4+-sqrt(4*2*(-1)))/(2*2)

= (4+-2i*sqrt(2))/(2*2)

= (2*(2+-i*sqrt(2)))/(2*2)

= (2+-i*sqrt(2))/2

=1+-(i*sqrt(2))/2)