Solve (3-i)(2-i)-2i^2-sqrt{-9}

$$( 3 - i ) ( 2 - i ) - 2 i ^ { 2 } - \sqrt { - 9 }$$

Evaluate

$$7-8i$$

Solution Steps

Multiply 3-i and 2-i to get 5-5i.

$$5-5i-2i^{2}-\sqrt{-9}$$

Calculate i to the power of 2 and get -1.

$$5-5i-2\left(-1\right)-\sqrt{-9}$$

Multiply 2 and -1 to get -2.

$$5-5i-\left(-2\right)-\sqrt{-9}$$

The opposite of -2 is 2.

$$5-5i+2-\sqrt{-9}$$

Add 5-5i and 2 to get 7-5i.

$$7-5i-\sqrt{-9}$$

Calculate the square root of -9 and get 3i.

$$7-5i-3i$$

Subtract 3i from 7-5i to get 7-8i.

$$7-8i$$

Real Part

$$7$$

Solution Steps

Multiply 3-i and 2-i to get 5-5i.

$$Re(5-5i-2i^{2}-\sqrt{-9})$$

Calculate i to the power of 2 and get -1.

$$Re(5-5i-2\left(-1\right)-\sqrt{-9})$$

Multiply 2 and -1 to get -2.

$$Re(5-5i-\left(-2\right)-\sqrt{-9})$$

The opposite of -2 is 2.

$$Re(5-5i+2-\sqrt{-9})$$

Add 5-5i and 2 to get 7-5i.

$$Re(7-5i-\sqrt{-9})$$

Calculate the square root of -9 and get 3i.

$$Re(7-5i-3i)$$

Subtract 3i from 7-5i to get 7-8i.

$$Re(7-8i)$$

The real part of 7-8i is 7.

$$7$$