4(y-7) - 2y(1-3) =6y^2 solve?

Question

Im really bad at this, please show me and explain all the steps to get the answer. thanks!!!!!!!!!!!

Answer 1

Assuming question is meant to be:-
4(y - 7) - 2y(y - 3) = 6y²
4y - 28 - 2y² + 6y = 6y²
8y² - 10y + 28 = 0
4y² - 5y + 14 =0
y = [ 5 ± √(25 - 224) ] / 8
y = [ 5 ± √(- 199) ] / 8
y = [ 5 ± i √(199) ] / 8

Answer 2

4(y - 7) - 2y(1 - 3) = 6y² >>>Expand the bracket
4y - 28 - 2y + 6 - 6y² = 0
-6y² + 4y - 2y - 28 + 6 = 0
-6y² + 2y - 22 = 0 >>>Make in general form
You can using completing the square or quadratic formula. You can't using trinomial method or cross multiplication method because the solution cannot be determined.

Completing the square
-6y² + 2y - 22 = 0
-6y² + 2y = 22 >>>Rearrange in the form x² + px = q
-6y² + 2y + (2 / 2)² = 22 + (2 / 2)² >>>Add (coefficient of x / 2)² to both side. Coefficient of x is q which is 2.
-6y² + 2y + 1² = 22 + 1²
(-6y + 1)² = 22
√(-6y + 1)² = √22
-6y + 1 = √22
-6y = -1 ± √22

-6y = -1 + √22
-6y = 3.690
y = 3.690 / -6
y = -0.615 >>>3 decimal point

-6y = -1 - √22
-6y = -5.690
y = -5.690 / -6
y = 0.948 >>>3 decimal point

Thus y = -0.615 or y = 0.948

Answer 3

open the brackets

4y -28 -2y +6y -6y^2 = 6y^2

8y -28 = 6y^2

bringing 6y^2 to other side
6y^2 - 8y + 28=0
divide the equation by 2
3y^2 -4y +14 =0

use the discriminant formula

and get the final answer it will have imaginary roots

Answer 4

if you simplify that it gets to

6y^2 + 4y + 28 = 0

and divide both sides by 2 to get

3y^2 + 2y + 14 = 0

and use the quadratic formula to get the value of y

but i think that leaves you with an imaginary answer

Answer 5

4(y-7)-2y(1-3)=6y^2
4y-28-2y(-2)=6y^2
6y^2-8y+28=0
3y^2-4y+14=0
delta=(16-4*3*14)^.5<0
then
this equation has not real answer.

Answer 6

ஐClueLessஐ

I Hate Math, Sorry I Couldn't Help You!
Good Luck!

Answer 7

Do some multiplying...maybe some division...possibly some subtraction and THERE YA GO.