Solve 2^2+sqrt{16}-3sqrt{-8}

$$2 ^ { 2 } + \sqrt { 16 } - 3 \sqrt { - 8 }$$

Evaluate (complex solution)

$$-6\sqrt{2}i+8\approx 8-8.485281374i$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$4+\sqrt{16}-3\sqrt{-8}$$

Calculate the square root of 16 and get 4.

$$4+4-3\sqrt{-8}$$

Add 4 and 4 to get 8.

$$8-3\sqrt{-8}$$

Factor -8=\left(2i\right)^{2}\times 2. Rewrite the square root of the product \sqrt{\left(2i\right)^{2}\times 2} as the product of square roots \sqrt{\left(2i\right)^{2}}\sqrt{2}. Take the square root of \left(2i\right)^{2}.

$$8-3\times \left(2i\right)\sqrt{2}$$

Multiply -3 and 2i to get -6i.

$$8-6i\sqrt{2}$$

Real Part (complex solution)

$$8$$

Evaluate

$$\text{Indeterminate}$$