Answer:
3y^2 - (y + 2) (y - 2) = 0
<=> 3y^2 - (y^2 - 4) = 0
<=> 2y^2 + 4 =0
<=>y^2 + 2 = 0
=> Because y^2 is always equal or larger than 0, there is no real solution.
Hope this helps!
:)
[tex]2 {y}^{2} + 4[/tex]
Step-by-step explanation:
[tex]3 {y}^{2} - (y + 2)(y - 2) \\ 3 {y}^{2} - ( {y}^{2} - 4) \\ 3 {y}^{2} - {y}^{2} + 4 \\ = 2 {y}^{2} + 4 \\ = 2( {y}^{2} + 2)[/tex]
