Solve {left(sqrt{32-3}right)}^2+6sqrt{32}

$${ \left( \sqrt{ 32-3 } \right) }^{ 2 } +6 \sqrt{ 32 }$$

Evaluate

$$24\sqrt{2}+29\approx 62.941125497$$

Solution Steps

Subtract 3 from 32 to get 29.

$$\left(\sqrt{29}\right)^{2}+6\sqrt{32}$$

The square of \sqrt{29} is 29.

$$29+6\sqrt{32}$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$29+6\times 4\sqrt{2}$$

Multiply 6 and 4 to get 24.

$$29+24\sqrt{2}$$