V question for ya?

////9,-9

3y^2 - 243 = 0
(3y^2 - 243)/3 = 0/3
y^2 - 81 = 0
y^2 + 9y - 9y - 81 = 0
(y^2 + 9y) - (9y + 81) = 0
y(y + 9) - 9(y + 9) = 0
(y + 9)(y - 9) = 0

y + 9 = 0
y = -9

y - 9 = 0
y = 9

∴ y = ±9

... Is that:
3(y^2) - 243 = 0
?
(with the up arrow signifying 'to the power of', and the brackets showing more definitely what is put to the power of)

If so, then first, add 243 to both sides:

3(y^2) - 243 + 243 = 0 + 243

3(y^2) = 243

Then, divide both sides by three:

(3(y^2))/3 = 243/3

y^2 = 81

Then, find the positive and negative square root of the right hand side for y:

y = + or - √81

y = 9 or -9

Which is the last possible answer that you supplied.

3y^2-243=0

3y^2=243

y^2=81

y is -9 or 9

If you are confused by the square root of 81 ....don't be...yes, in the mathematical community the square root of a number is defined by the positive principal square root. But you are asking what numbers actually raise to the second power equal 81 and that is -9 and 9.

3y² – 243 = 0
3(y² – 81) = 0
(y² – 81) = 0
y² = 81
y √81 = ± 9
-----------

3(9 + y)(-9 + y)
set them equal to zero

3y^2-243=0
ory^2-81=0 [dividing by 3 on both sides]
or y^2-9^2=0
or(y+9)(y-9)=0
or y=-9 or +9 is the correct answer

3 y 2 = 243
y 2 = 243/3=81
y= square root of 81.
so y= 9 , y=-9.

y=9, -9

y² = 81
y = ± 9