Help how would I solve this equation: 2z^2 = 2(z + 3)^2

Expanding (z + 3)², we obtain:

2z² = 2(z² + 6z + 9)

Divide both sides by 2 to get:

z² = z² + 6z + 9

Finally, rearrange the terms:

z² - z² = z² - z² + 6z + 9
0 = 6z + 9 [Subtract both sides by z²]
-6z = 6z - 6z + 9
-6z = 9 [Subtract both sides by 6z]
z = -9/6 [Divide both sides by -6]

So z = -3/2

I hope this helps!

2z^2 = 2(z + 3)^2

z^2 = (z + 3)^2

the absolute value of z = the absolute value of z + 3

by means of trial and error you get that z=-1.5'


as the absolute value of -1.5= 1.5

2z^2 = 2(z + 3)^2

z^2 = (z +3)^2
z^2 = z^2 +6z +9
6z +9 = 0
6z = -9
z = -9/6 = -3/2 = - 1 1/2