How do you complete the square with a fraction?

Question

I've been given the question 2x^2 + 6x =5. Now, this is the first time I've tried to get rid of the coefficient, but the answer I'm getting isn't the same as in the answers. First, I get rid of the first coefficent, making it x^2 +3x =5/2. How would I go about turning that 3x into something I could use to square? And then how would I follow through with the rest of this? Thanks for the help! ^-^

Answer 1

x^2 + 3x = 5/2

The rule is : take half of 3 (= 3/2) and then square it (= 9/4).

Now add this 9/4 to both sides.

x^2 + 3x + 9/4 = 5/2 + 9/4

(x + 3/2)^2 = 19/4

Take the square root of both sides :
x + 3/2 = ± 19/4

Subtract 3/2 from both sides :
x = ± 19/4 - 3/2

Thus x = 13/4 or -25/4.

Answer 2

2x² + 6x = 5
x² + 3x = 5/2
x² + 3x + 9/4 = 5/2 + 9/4
(x + 3/2)² = 19/4
x + 3/2 = (±√19) / 2
x = (- 3/ 2) ± (√19 ) / 2
x = (1/2) (- 3 ±√19)

Answer 3

first,square half of the coefficient of x ie (3/2)^2=9/4.
Add this to both sides,
x^2+3x+9/4=5/2+9/4
(x+3/2)^2=19/4
x+3/2=+or-sqr19/4
x=-3/2+or-sqr19/4

Answer 4

multiply by 2
4x^2+12x =10
...

otherwise also
x^2+2*(3/2)*x = 5/2
....

Answer 5

use the quadradic formula

-b +/- sq.root of (-b^2-4(a)(c))
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2(a)

-6 +/- sq.root of ((-6)^2-4(2)(-5))
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2(2)