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\left(5w-2\right)\left(5w+2\right)=0
Consider 25w^{2}-4. Rewrite 25w^{2}-4 as \left(5w\right)^{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).
w=\frac{2}{5} w=-\frac{2}{5}
To find equation solutions, solve 5w-2=0 and 5w+2=0.
25w^{2}=4
Add 4 to both sides. Anything plus zero gives itself.
w^{2}=\frac{4}{25}
Divide both sides by 25.
w=\frac{2}{5} w=-\frac{2}{5}
Take the square root of both sides of the equation.
25w^{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.
w=\frac{0±\sqrt{0^{2}-4\times 25\left(-4\right)}}{2\times 25}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
w=\frac{0±\sqrt{-4\times 25\left(-4\right)}}{2\times 25}
Square 0.
w=\frac{0±\sqrt{-100\left(-4\right)}}{2\times 25}
Multiply -4 times 25.
w=\frac{0±\sqrt{400}}{2\times 25}
Multiply -100 times -4.
w=\frac{0±20}{2\times 25}
Take the square root of 400.
w=\frac{0±20}{50}
Multiply 2 times 25.
w=\frac{2}{5}
Now solve the equation w=\frac{0±20}{50} when ± is plus. Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.
w=-\frac{2}{5}
Now solve the equation w=\frac{0±20}{50} when ± is minus. Reduce the fraction \frac{-20}{50} to lowest terms by extracting and canceling out 10.
w=\frac{2}{5} w=-\frac{2}{5}
The equation is now solved.