Solve asqrt{b}x+(b^2-ax)x-bc=0

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

Solve for a

$$\left\{\begin{matrix}a=-\frac{b\left(bx-c\right)}{x\left(-x+\sqrt{b}\right)}\text{, }&x\neq 0\text{ and }\left(x<0\text{ or }b\neq x^{2}\right)\text{ and }b\geq 0\\a\in \mathrm{R}\text{, }&\left(c=0\text{ and }x=0\text{ and }b>0\right)\text{ or }\left(b=c^{\frac{2}{3}}\text{ and }x=\sqrt[3]{c}\text{ and }c>0\right)\text{ or }\left(x=0\text{ and }b=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a\sqrt{b}x+b^{2}x-ax^{2}-bc=0$$

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

$$a\sqrt{b}x-ax^{2}-bc=-b^{2}x$$

Add bc to both sides.

$$a\sqrt{b}x-ax^{2}=-b^{2}x+bc$$

Reorder the terms.

$$\sqrt{b}ax-ax^{2}=-xb^{2}+bc$$

Combine all terms containing a.

$$\left(\sqrt{b}x-x^{2}\right)a=-xb^{2}+bc$$

The equation is in standard form.

$$\left(\sqrt{b}x-x^{2}\right)a=bc-xb^{2}$$

Divide both sides by \sqrt{b}x-x^{2}.

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

Dividing by \sqrt{b}x-x^{2} undoes the multiplication by \sqrt{b}x-x^{2}.

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

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

$$a=\frac{b\left(c-bx\right)}{x\left(-x+\sqrt{b}\right)}$$