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y^{2}-4=0
Subtract 4 from both sides.
\left(y-2\right)\left(y+2\right)=0
Consider y^{2}-4. Rewrite y^{2}-4 as y^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
y=2 y=-2
To find equation solutions, solve y-2=0 and y+2=0.
y=2 y=-2
Take the square root of both sides of the equation.
y^{2}-4=0
Subtract 4 from both sides.
y=\frac{0±\sqrt{0^{2}-4\left(-4\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{0±\sqrt{-4\left(-4\right)}}{2}
Square 0.
y=\frac{0±\sqrt{16}}{2}
Multiply -4 times -4.
y=\frac{0±4}{2}
Take the square root of 16.
y=2
Now solve the equation y=\frac{0±4}{2} when ± is plus. Divide 4 by 2.
y=-2
Now solve the equation y=\frac{0±4}{2} when ± is minus. Divide -4 by 2.
y=2 y=-2
The equation is now solved.