Solve for y

Question

4y²-5=12y

Answer 1

Hi Lila;

4y²-5=12y

Let's have everything on one side...

4y²-12y-5=0

(2y-3)(2y-3)

Let's FOIL...

FIRST...(2y)(2y)=4y²

OUTER...(2y)(-3)=-6y

INNER...(-3)(2y)=-6y

LAST...(-3)(-3)=9

4y²-6y-6y+9

4y²-12y+9

Obviously, the constant we want is NOT +9.  We want -5...

-5=9-14

4y²-12y+9-14=0

(2y-3)²-14=0

(2y-3)²=14

Square-root both sides...

2y-3=+√14.......2y-3=-√14

2y=3+√14........2y=3-√14

y=(3+√14)/2....y=(3-√14)/2

Answer 2

4y^2 - 5 = 12y

 

4y^2 -12y - 5 = 0

 

(2y -5)(2y -1 ) = 0

 

2y - 5 = 0

2y = 5

y = 5/2

 

and

 

2y -1 =0

2y = 1

y = 1/2