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x^{2}-4.41=0
Subtract 4.41 from both sides.
\left(x-\frac{21}{10}\right)\left(x+\frac{21}{10}\right)=0
Consider x^{2}-4.41. Rewrite x^{2}-4.41 as x^{2}-\left(\frac{21}{10}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
x=\frac{21}{10} x=-\frac{21}{10}
To find equation solutions, solve x-\frac{21}{10}=0 and x+\frac{21}{10}=0.
x=\frac{21}{10} x=-\frac{21}{10}
Take the square root of both sides of the equation.
x^{2}-4.41=0
Subtract 4.41 from both sides.
x=\frac{0±\sqrt{0^{2}-4\left(-4.41\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -4.41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\left(-4.41\right)}}{2}
Square 0.
x=\frac{0±\sqrt{17.64}}{2}
Multiply -4 times -4.41.
x=\frac{0±\frac{21}{5}}{2}
Take the square root of 17.64.
x=\frac{21}{10}
Now solve the equation x=\frac{0±\frac{21}{5}}{2} when ± is plus.
x=-\frac{21}{10}
Now solve the equation x=\frac{0±\frac{21}{5}}{2} when ± is minus.
x=\frac{21}{10} x=-\frac{21}{10}
The equation is now solved.