Solve b^2x^2+b(a-c)x-ac=0

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

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=-bx\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&c=bx\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply b by a-c.

$$b^{2}x^{2}+\left(ba-bc\right)x-ac=0$$

Use the distributive property to multiply ba-bc by x.

$$b^{2}x^{2}+bax-bcx-ac=0$$

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

$$bax-bcx-ac=-b^{2}x^{2}$$

Add bcx to both sides.

$$bax-ac=-b^{2}x^{2}+bcx$$

Reorder the terms.

$$abx-ac=-b^{2}x^{2}+bcx$$

Combine all terms containing a.

$$\left(bx-c\right)a=-b^{2}x^{2}+bcx$$

The equation is in standard form.

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

Divide both sides by bx-c.

$$\frac{\left(bx-c\right)a}{bx-c}=\frac{bx\left(c-bx\right)}{bx-c}$$

Dividing by bx-c undoes the multiplication by bx-c.

$$a=\frac{bx\left(c-bx\right)}{bx-c}$$

Divide bx\left(-bx+c\right) by bx-c.

$$a=-bx$$

Solve for a

$$\left\{\begin{matrix}\\a=-bx\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&c=bx\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply b by a-c.

$$b^{2}x^{2}+\left(ba-bc\right)x-ac=0$$

Use the distributive property to multiply ba-bc by x.

$$b^{2}x^{2}+bax-bcx-ac=0$$

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

$$bax-bcx-ac=-b^{2}x^{2}$$

Add bcx to both sides.

$$bax-ac=-b^{2}x^{2}+bcx$$

Reorder the terms.

$$abx-ac=-b^{2}x^{2}+bcx$$

Combine all terms containing a.

$$\left(bx-c\right)a=-b^{2}x^{2}+bcx$$

The equation is in standard form.

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

Divide both sides by bx-c.

$$\frac{\left(bx-c\right)a}{bx-c}=\frac{bx\left(c-bx\right)}{bx-c}$$

Dividing by bx-c undoes the multiplication by bx-c.

$$a=\frac{bx\left(c-bx\right)}{bx-c}$$

Divide bx\left(-bx+c\right) by bx-c.

$$a=-bx$$

Solve for b (complex solution)

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

Solve for b

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