Solve abx^2+(b^2-ac)x-bc=0

$$a b x ^ { 2 } + ( b ^ { 2 } - a c ) x - b c = 0$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{b}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&c=bx\text{ or }\left(b=0\text{ and }x=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply b^{2}-ac by x.

$$abx^{2}+b^{2}x-acx-bc=0$$

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

$$abx^{2}-acx-bc=-b^{2}x$$

Add bc to both sides.

$$abx^{2}-acx=-b^{2}x+bc$$

Reorder the terms.

$$abx^{2}-acx=-xb^{2}+bc$$

Combine all terms containing a.

$$\left(bx^{2}-cx\right)a=-xb^{2}+bc$$

The equation is in standard form.

$$\left(bx^{2}-cx\right)a=bc-xb^{2}$$

Divide both sides by bx^{2}-cx.

$$\frac{\left(bx^{2}-cx\right)a}{bx^{2}-cx}=\frac{b\left(c-bx\right)}{bx^{2}-cx}$$

Dividing by bx^{2}-cx undoes the multiplication by bx^{2}-cx.

$$a=\frac{b\left(c-bx\right)}{bx^{2}-cx}$$

Divide b\left(-xb+c\right) by bx^{2}-cx.

$$a=-\frac{b}{x}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{b}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&c=bx\text{ or }\left(b=0\text{ and }x=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply b^{2}-ac by x.

$$abx^{2}+b^{2}x-acx-bc=0$$

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

$$abx^{2}-acx-bc=-b^{2}x$$

Add bc to both sides.

$$abx^{2}-acx=-b^{2}x+bc$$

Reorder the terms.

$$abx^{2}-acx=-xb^{2}+bc$$

Combine all terms containing a.

$$\left(bx^{2}-cx\right)a=-xb^{2}+bc$$

The equation is in standard form.

$$\left(bx^{2}-cx\right)a=bc-xb^{2}$$

Divide both sides by bx^{2}-cx.

$$\frac{\left(bx^{2}-cx\right)a}{bx^{2}-cx}=\frac{b\left(c-bx\right)}{bx^{2}-cx}$$

Dividing by bx^{2}-cx undoes the multiplication by bx^{2}-cx.

$$a=\frac{b\left(c-bx\right)}{bx^{2}-cx}$$

Divide b\left(-xb+c\right) by bx^{2}-cx.

$$a=-\frac{b}{x}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=-ax\text{, }&\text{unconditionally}\\b=\frac{c}{x}\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&c=0\text{ and }x=0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}\\b=-ax\text{, }&\text{unconditionally}\\b=\frac{c}{x}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&c=0\text{ and }x=0\end{matrix}\right.$$