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\sqrt{\frac{\left(1.69-0.7^{2}+3.2\right)^{2}}{11^{2}}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Calculate 1.3 to the power of 2 and get 1.69.
\sqrt{\frac{\left(1.69-0.49+3.2\right)^{2}}{11^{2}}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Calculate 0.7 to the power of 2 and get 0.49.
\sqrt{\frac{\left(1.2+3.2\right)^{2}}{11^{2}}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Subtract 0.49 from 1.69 to get 1.2.
\sqrt{\frac{4.4^{2}}{11^{2}}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Add 1.2 and 3.2 to get 4.4.
\sqrt{\frac{19.36}{11^{2}}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Calculate 4.4 to the power of 2 and get 19.36.
\sqrt{\frac{19.36}{121}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Calculate 11 to the power of 2 and get 121.
\sqrt{\frac{1936}{12100}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Expand \frac{19.36}{121} by multiplying both numerator and the denominator by 100.
\sqrt{\frac{4}{25}\times 5+\left(2.8^{2}-0.23\right)\times 10^{2}}
Reduce the fraction \frac{1936}{12100} to lowest terms by extracting and canceling out 484.
\sqrt{\frac{4}{5}+\left(2.8^{2}-0.23\right)\times 10^{2}}
Multiply \frac{4}{25} and 5 to get \frac{4}{5}.
\sqrt{\frac{4}{5}+\left(7.84-0.23\right)\times 10^{2}}
Calculate 2.8 to the power of 2 and get 7.84.
\sqrt{\frac{4}{5}+7.61\times 10^{2}}
Subtract 0.23 from 7.84 to get 7.61.
\sqrt{\frac{4}{5}+7.61\times 100}
Calculate 10 to the power of 2 and get 100.
\sqrt{\frac{4}{5}+761}
Multiply 7.61 and 100 to get 761.
\sqrt{\frac{3809}{5}}
Add \frac{4}{5} and 761 to get \frac{3809}{5}.
\frac{\sqrt{3809}}{\sqrt{5}}
Rewrite the square root of the division \sqrt{\frac{3809}{5}} as the division of square roots \frac{\sqrt{3809}}{\sqrt{5}}.
\frac{\sqrt{3809}\sqrt{5}}{\left(\sqrt{5}\right)^{2}}
Rationalize the denominator of \frac{\sqrt{3809}}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.
\frac{\sqrt{3809}\sqrt{5}}{5}
The square of \sqrt{5} is 5.
\frac{\sqrt{19045}}{5}
To multiply \sqrt{3809} and \sqrt{5}, multiply the numbers under the square root.