Can anybody tell me how to solve?

Question

2y^2-3y=1

Answer 1

2y^2-3y-1=0

(2y+1)(y-1)=0

2y+1=0
y-1=0

solve for y

Answer 2

2y^2-3y=1

First subtract 1 from both sides

2y^2-3y -1

This can be then factorised by using the quadratic equation

y = (3 + or - sqrt(3^2 + 4*2*1))/ (2*2)

y = ( 3 + or - sqrt(9 + 8))/ 4

y = (3 + or - sqrt(17))/ 4

y1= (3 + sqrt(17))/ 4 = 1.7808

y2 = (3 - sqrt(17))/ 4 = -0.2808.

Answer 3

divide the whole thing by 2
y^2 - 1.5y = 0.5
Add the square of half of 1.5 to both sides

y^2 - (3/2)y + 9/16 = 8/16 + 9/16 = 17/16

which gives (y - 3/4)^2 = 17/16
y - 3/4 = +/-sqrt(17/16)
y = 3/4 +-sqrt(17/16)

Answer 4

Hi,

Move everything to one side.

2y^2 -3y-1=0

This does not factor, so plug it into the quadratic formula. The Quadratic Formula uses the "a", "b", and "c" from "ax^2 + bx + c", where "a", "b", and "c" are just numbers. For you, a = 2, b = -3, and c = -1.

[ -b + or - sqrt(b^2 -4ac)]/(2a) is the quadratic formula. For this problem, it becomes:

y = [ -b + or - sqrt(b^2 -4ac)]/(2a)

y = [ -(-3) + or - sqrt((-3)^2 -4(2)(-1))]/(2(2))

OR

-(-3) + or - sqrt((-3)^2 -4(2)(-1))]
-----------------------------------------------
________ (2(2))


which simplifies to

3 + or - sqrt(9 +8)
------------------------- = y
4

3 + or - sqrt(17)
---------------------- = y
4

This is your answer unless you want decimals for it. They are about 1.78 or -.28

I hope that helps!

Answer 5

Sure! Subtract the 1.

You get 2y^2-3y-1=0

Use the quadratic formula: -b +/- b^2 - 4ac
all over 2a. (It looks better on paper.)

Simplify, and you get 3 +/- 17 all over 4.

Good luck.

Answer 6

2y² - 3y - 1 = 0
y = [3 ± √(9 + 8)] / 4
y = [3 ± √17] / 4
y = 0.75 ± 1.03
y = 1.78, - 0 .28

Answer 7

move the one to the other side and factor so you have two smaller equations of something times y plus or minus something, then you set both of those equal to 0 and solve for y, you should have 2 different answers for y, put them both down as y = __ or __

Answer 8

Move the 1 over to the other side and factor.

Answer 9

2y^2-3y=1
2y^2-3y-1=0
(2y+1)(y-1)
y=1/2 (or) y=1





PcD

Answer 10

hehe no.