How do you solve y(3y-4)= 7?

Question

How do you solve y(3y-4)= 7
I've got the answer by trying different numbers but i want to know how
you would get the answers i.e what method would you use.

Answer 1

y(3y-4)= 7
3y^2-4y = 7
3y^2-4y-7=0
[3y-7][y+1] = 0
3y - 7 = 0
y = 7/3
y + 1 = 0
y = -1
So y = (7/3) and -1

Answer 2

There are several ways to solve this. We will use the quadratic equation. A quadratic equation looks like ax**2 + bx + c = 0 (read that as a x squared plus b x plus c equals 0) where x is the variable, and a,b,c are constants. Since this is a second degree equation (because of x**2 term), there will two possible values for x that will satisfy. They are

(-b + (b**2 - 4ac))/2a ---- Value 1

and (-b - (b**2 - 4ac))/2a ---- Value 2

So, now let us take your problem and convert it to look like a quadratic. Expanding the left side, the equation becomes
3y**2 - 4y = 7. If we add -7 to the right, the right side will become 0 which is what we need to make it to like the quadratic equation above. However, this is an equation. Since we added -7 to the right, we need to do the same to the left. So the equation becomes,

3y**2 - 4y -7 = 0 and that looks like the quadratic equation, doesn't it? Simply, instead of the variable x we have the variable y. The constants are a=3, b = -4 and c = -7. Using the formulas I gave earlier in values 1 and 2, and plugging in for a, b, and c, we get the following two solutions for your problem.

2 1/3 and -1

Hope that helps.

Answer 3

It is a quadratic equation
3y²-4y-7 = 0
D = 16 +84 = 100 = 10²

so the solutions are y1 = (4+10)/6 = 7/3
y2 =(4-10)/6 = -1

Answer 4

y(3y - 4) = 7
3y^2 - 4y = 7
3y^2 - 4y - 7 = 0
(3y - 7)(y + 1) = 0 {factor the trinomial}

y = 7/3, -1

Answer 5

y (3y - 4) = 7

multiply out

3y^2 - 4y = 7

subtract 7

3y^2 - 4y - 7 = 0

find the square? (i forget what it is called)

(3y - 7) (y + 1) = 0

y = -1, 7/3

Answer 6

y(3y-4)=7
3y^2-4y-7=0
divide by 3
y^2-(4/3)y-(7/3)=0
factorize
(y-7/3)(y+1)=0

>>>>y= 7/3 or y= -1

i hope that this helps

Answer 7

3y^2 -4y -7=0
use the formula y= (-b +/-square root(b^2-4ac))/2a where a=3, b=4, c=7.
so, y= (4+/-10)6
y=-1 or y=14/6

Answer 8

Solutions: y=2.33333333333333,y=-1.

Text form: y*(3y-4)=7 simplifies to 0=0

Answer 9

3y^2 - 4y - 7 = 0
Use the quadratic formula.

y1 = 7/3
y2 = -1

Answer 10

y (3y - 4) = 7
3y² - 4y = 7
3y² - 4y - 7 = 0

Just to prove the above answers, bring in the quadratic equation:
y = - b ± √(b² - 4ac)
_________________
2a

- (-4) ± √(-4)² - 4 (3)(-7)
______________________
2 (3)

- 4 ± √16 + 84
_______________
6

- 4 ± √100
_________
6

- 4 ± 10
________
6

Answer 1: 1
Answer 2: 14/6 = 7/3