^{2}-5=12y

^{2}-5=12y

^{2}-12y-5=0

^{2}

^{2}-6y-6y+9

^{2}-12y+9

^{2}-12y+9-14=0

^{2}-14=0

^{2}=14

4y^{2}-5=12y

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Hi Lila;

4y^{2}-5=12y

Let's have everything on one side...

4y^{2}-12y-5=0

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

Let's FOIL...

FIRST...(2y)(2y)=4y^{2}

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

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

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

4y^{2}-6y-6y+9

4y^{2}-12y+9

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

-5=9-14

4y^{2}-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

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

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