Solve kz=(2+i)m^2-3(i+1)m-2(1-i)

$$k z = ( 2 + i ) m ^ { 2 } - 3 ( i + 1 ) m - 2 ( 1 - i )$$

Solve for k

$$\left\{\begin{matrix}k=-\frac{\left(-2-i\right)m^{2}+\left(3+3i\right)m+\left(2-2i\right)}{z}\text{, }&z\neq 0\\k\in \mathrm{C}\text{, }&\left(m=2\text{ or }m=-\frac{1}{5}+\frac{3}{5}i\right)\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 2 and 1-i to get 2-2i.

$$kz=\left(2+i\right)m^{2}-3\left(i+1\right)m-\left(2-2i\right)$$

Use the distributive property to multiply -3 by i+1.

$$kz=\left(2+i\right)m^{2}+\left(-3i-3\right)m-\left(2-2i\right)$$

Use the distributive property to multiply -3i-3 by m.

$$kz=\left(2+i\right)m^{2}+\left(-3-3i\right)m-\left(2-2i\right)$$

Multiply -1 and 2-2i to get -2+2i.

$$kz=\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)$$

The equation is in standard form.

$$zk=\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)$$

Divide both sides by z.

$$\frac{zk}{z}=\frac{\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)}{z}$$

Dividing by z undoes the multiplication by z.

$$k=\frac{\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)}{z}$$

Solve for m

$$m=\left(\frac{1}{5}-\frac{1}{10}i\right)\sqrt{\left(8+4i\right)kz+\left(24+10i\right)}+\left(\frac{9}{10}+\frac{3}{10}i\right)$$

$$m=\left(-\frac{1}{5}+\frac{1}{10}i\right)\sqrt{\left(8+4i\right)kz+\left(24+10i\right)}+\left(\frac{9}{10}+\frac{3}{10}i\right)$$