Solve z^2+(p+5i)z+q(2-i)=0quadx

$$z ^ { 2 } + ( p + 5 i ) z + q ( 2 - i ) = 0 \quad x$$

Solve for p

$$\left\{\begin{matrix}p=-z+\frac{\left(-2+i\right)q}{z}-5i\text{, }&z\neq 0\\p\in \mathrm{C}\text{, }&q=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply p+5i by z.

$$z^{2}+pz+5iz+q\left(2-i\right)=0x$$

Anything times zero gives zero.

$$z^{2}+pz+5iz+q\left(2-i\right)=0$$

Subtract z^{2} from both sides. Anything subtracted from zero gives its negation.

$$pz+5iz+q\left(2-i\right)=-z^{2}$$

Subtract 5iz from both sides.

$$pz+q\left(2-i\right)=-z^{2}-5iz$$

Subtract q\left(2-i\right) from both sides.

$$pz=-z^{2}-5iz-q\left(2-i\right)$$

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

$$pz=-z^{2}-5iz+\left(-2+i\right)q$$

The equation is in standard form.

$$zp=\left(-2+i\right)q-5iz-z^{2}$$

Divide both sides by z.

$$\frac{zp}{z}=\frac{\left(-2+i\right)q-5iz-z^{2}}{z}$$

Dividing by z undoes the multiplication by z.

$$p=\frac{\left(-2+i\right)q-5iz-z^{2}}{z}$$

Divide -z^{2}-5iz+\left(-2+i\right)q by z.

$$p=-z+\frac{\left(-2+i\right)q}{z}-5i$$

Solve for q

$$q=\left(-\frac{2}{5}-\frac{1}{5}i\right)z\left(z+p+5i\right)$$

Steps for Solving Linear Equation

Use the distributive property to multiply p+5i by z.

$$z^{2}+pz+5iz+q\left(2-i\right)=0x$$

Anything times zero gives zero.

$$z^{2}+pz+5iz+q\left(2-i\right)=0$$

Subtract z^{2} from both sides. Anything subtracted from zero gives its negation.

$$pz+5iz+q\left(2-i\right)=-z^{2}$$

Subtract pz from both sides.

$$5iz+q\left(2-i\right)=-z^{2}-pz$$

Subtract 5iz from both sides.

$$q\left(2-i\right)=-z^{2}-pz-5iz$$

The equation is in standard form.

$$\left(2-i\right)q=-z^{2}-pz-5iz$$

Divide both sides by 2-i.

$$\frac{\left(2-i\right)q}{2-i}=-\frac{z\left(z+p+5i\right)}{2-i}$$

Dividing by 2-i undoes the multiplication by 2-i.

$$q=-\frac{z\left(z+p+5i\right)}{2-i}$$

Divide -z\left(z+p+5i\right) by 2-i.

$$q=\left(-\frac{2}{5}-\frac{1}{5}i\right)z\left(z+p+5i\right)$$