Solve 3+17^2-(5-i)z+2*2i=0

$$3 + 17 ^ { 2 } - ( 5 - i ) z + 2 \cdot 2 i = 0$$

Solve for z

$$z=56+12i$$

Steps for Solving Linear Equation

Calculate 17 to the power of 2 and get 289.

$$3+289-\left(5-i\right)z+2\times \left(2i\right)=0$$

Add 3 and 289 to get 292.

$$292-\left(5-i\right)z+2\times \left(2i\right)=0$$

Multiply 2 and 2i to get 4i.

$$292-\left(5-i\right)z+4i=0$$

Subtract 4i from both sides. Anything subtracted from zero gives its negation.

$$292-\left(5-i\right)z=-4i$$

Multiply -1 and 5-i to get -5+i.

$$292+\left(-5+i\right)z=-4i$$

Subtract 292 from both sides.

$$\left(-5+i\right)z=-4i-292$$

The equation is in standard form.

$$\left(-5+i\right)z=-292-4i$$

Divide both sides by -5+i.

$$\frac{\left(-5+i\right)z}{-5+i}=\frac{-292-4i}{-5+i}$$

Dividing by -5+i undoes the multiplication by -5+i.

$$z=\frac{-292-4i}{-5+i}$$

Divide -292-4i by -5+i.

$$z=56+12i$$