quadratic equation , how do I solve using the complete the square method?

Question

2n^2= -1+6n

Answer 1

write the expression as (something)^2 = number, next somtheing is + or - sqrt(number)


2n^2= -1+6n

2n^2 - 6n + 1 = 0

[ sqrt(2) * N - sqrt(2)*6/4 ] ^2 -18 + 1 = 0;
thus
[ sqrt(2) * N - sqrt(2)*6/4 ] ^2 = 17

sqrt(2) * N - sqrt(2)*6/4 = + sqrt(17) or - sqrt(17)

etc ..

Answer 2

2n^2= -1+6n
=> 2n^2 - 6n + 1 = 0
=> n^2 - 3n + 1/2 = 0
=> (n - 3/2)^2 - 9/4 + 2/4 =0
=> (n - 3/2)^2 = (√7/2)^2
=> n - 3/2 = ± √7/2
=> n = (1/2) * ( 3 ± √7)

Answer 3

2n^2 = - 1 + 6n

2n^2 - 6n + 1= 0

divide by 2 to make x^2 coefficient 1

n^2 - 3n + 1/2 = 0

(n)^2 - 2(3/2) n + 1/2 = 0

so to make it perfect square (3/2)^2 = 9/4 is missing

add 9/4 both sides

n^2 - 3n + 9/4 + 1/2 = 9/4

(n-3/2)^2 = 9/4 - 1/2

(n - 3/2)^2 = 7/4

n - 3/2 = +/- sqrt(7/4)

n = 3/2 +/ - sqrt(7)/2

n = [3 + sqrt(7)]/2 or [3 - sqrt(7)]/2

Answer 4

2n^2-6n=-1
n^2-3n=-1/2

complete the square by adding 9/4 to each side we get

n^2-3n+9/4=9/4-1/2=7/4

(n-3/2)^2=7/4

n-3/2= +-sqrt7/2

n= +-sqrt7/2+3/2

Answer 5

2n^2-6n+1=0
n^2-3n+(1\2)=0

use (b^2-4ac) and then use the equation [(-b+/-b^2-4ac)/2a]

it comes out n=2.207 or n=0.7928

Answer 6

2n^2-6n+1=0
ax^2+bx+c=0
a=2 b=-6 c=1
x=[-b+ or - sqrt(b^2-4ac)]/2a
evaluate two x's using +sqrt(b^2-4ac) and -sqrt(b^2-4ac)
You'll get fractional answers.
x=2.82 x=0.177

Answer 7

2n^2 -6n = -1

2 (n^2 -3n) = -1

2 ((n - 1.5)^2 -2.25) = -1

(n -1.5)^2 - 2.25 = -0.5

(n - 1.5)^2 = 1.75

n - 1.5 = ±sqrt(7/4)

n = 1.5 ± (sqrt(7)/2)

n = 2.8228... or n = 0.17712...

Answer 8

2n^2-6n+1=0
n^2-3n+1/2=0
n^2-3n+(3/2)^2-(3/2)^2+1/2=0
(n-3/2)^2=(3/2)^2-1/2
(n-3/2)^2=9/4-1/2
(n-3/2)^2=7/4
n-3/2=sqrt(7)/2 or -sqrt(7)/2
so
n=3/2+sqrt(7)/2 or 3/2-sqrt(7)/2