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\left(R-2\right)\left(R+2\right)=0
Consider R^{2}-4. Rewrite R^{2}-4 as R^{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).
R=2 R=-2
To find equation solutions, solve R-2=0 and R+2=0.
R^{2}=4
Add 4 to both sides. Anything plus zero gives itself.
R=2 R=-2
Take the square root of both sides of the equation.
R^{2}-4=0
Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.
R=\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}.
R=\frac{0±\sqrt{-4\left(-4\right)}}{2}
Square 0.
R=\frac{0±\sqrt{16}}{2}
Multiply -4 times -4.
R=\frac{0±4}{2}
Take the square root of 16.
R=2
Now solve the equation R=\frac{0±4}{2} when ± is plus. Divide 4 by 2.
R=-2
Now solve the equation R=\frac{0±4}{2} when ± is minus. Divide -4 by 2.
R=2 R=-2
The equation is now solved.