Can someone explain how i work out y2=14 - 5y (the answer should be 2 but how do i work it out)?

Given Quadratic Equation in standard form :
y^2 +5y-14 =0
We can facorize the LHS expression by Method
of spilitting of Middle Term =>Their Product=(7y)x(-2y)= - 14
& Their Sum => (7y)+(--2y)= 5y
Hence, y^2+7y --2y--14 = 0
Or, y( y + 7) --2( y+ 7)= 0
Or, (y+7) (y-2)=0
By Zero Factor Theorem=> y+7=o=> y= -7
& y-2 = 0 => y=2

Thus,positive value of y=2 ,Negative value of y= --7
_________________________________________________Alternatively,Using Omar Khayyam's Quadratic Formula:

y= [ -b +( b^2-4ac)^ 1/2]/2a & y=[-b -(b^2-4ac) ^1/2/2a

a=1,b=5,c=-14=> y= [ -5+(25+56)^1/2]/2x1= 4^1/2=2
y= [-5 - ( 25 +56)^1/2]/2x1 =( -5-9)/2 = -7

Thus,postive value of y is 2 ,while negative value is --7

It depends what you meant to type

Option 1:

2y = 14 - 5y

First step: Add 5y to both sides

2y + 5y = 14 - 5y + 5y

Second step: Add up the Y's on each side

7y = 14

Third step: Divide both sides by 7

y = 2

Option 2:

y² = 14 - 5y

First Step: Add 5y to both sides

y² + 5y = 14 - 5y + 5y

Second Step: Add up the Y's on both sides

y² + 5y = 14

Third Step: Remove 14 from both sides

y² + 5y -14 = 14 -14

Fourth Step: Add up the numbers on both sides

y² + 5y -14 = 0

Fifth Step - We need to re-write the equation, with what is called FACTORING.
We need two numbers which add up to 5, and multiply to give -14. These are +7 and -2.
These give us the numbers in our brackets:

(y+7)(y-2) = 0

Sixth Step - we know one (or both) of the brackets needs to equal zero (to give an answer of 0).

Therefore, we have two equations:

Either y + 7 = 0
Or y - 2 = 0

Seventh Step - Solve the simple equations

If y + 7 = 0
Y = -7 (Subtract 7 from both sides)

If y -2 = 0
y = 2 (Add 2 to both sides)

So our answer is EITHER 2 or -7

Put into quadratic form, i.e. y^2 + 5y - 14 = 0 and solve using the quadratic formula y = -b +/- square root (b^2 -4ac) / 2a

y² + 5y - 14 = 0
(y + 7)(y - 2) = 0
y = - 7 , y = 2

y^2 = 14 - 5y
y^2 + 5y - 14 = 0
(y + 7)(y - 2) = 0
y = -7 or y = 2

its a quadratic equation
y^2+5y-14=0
general quadratic equation is ay^2+by+c=0
formula for the solution is given by y= [-b+(b^2-4ac)^(1/2)] / (2a)

a=1 b=5 c=-14
hence y= [-5+(25+56)^1/2] / 2=( -5+81^0.5)/2
but 81^0.5= + or - the square root of 81
hence y= (-5+9)/2 or (-5-9)/2
y= 2 or -7

Since it is a quadratic you will get two values of x

y^2+5y-14
=y^2+7y-2y-14
=y(y+7) -2(y+7)
=(y+7)(y-2)
=0
y is -7 or+2. If you are excluding -7, there must be another qualifying statement.

5y+2y=14
7y=14
y=2

U FACTOR THIS QUADRATIC EQUATION IN TO TWO BINOMIALS AND YOUR ANSWER MUST BE A POSITIVE ANSWER.

Y^2 =14 -5Y
Y^2 + 5Y-14 =0

(Y+ 7)(Y-2)= 0
Y= -7 OR Y=2 SINCE THE 2 IS POSITIVE THE ANSWER IS =2

y2=14-5y

adding 5y on both sides of the equation;

2y+5y=14-5y+5y

7y=14

y=14/7

y=2