Solve 2ax^2-ax+1=0

$$2 a x ^ { 2 } - a x + 1 = 0$$

Solve for a

$$a=-\frac{1}{x\left(2x-1\right)}$$

$$x\neq \frac{1}{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$2ax^{2}-ax=-1$$

Combine all terms containing a.

$$\left(2x^{2}-x\right)a=-1$$

Divide both sides by 2x^{2}-x.

$$\frac{\left(2x^{2}-x\right)a}{2x^{2}-x}=-\frac{1}{2x^{2}-x}$$

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

$$a=-\frac{1}{2x^{2}-x}$$

Divide -1 by 2x^{2}-x.

$$a=-\frac{1}{x\left(2x-1\right)}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{a\left(a-8\right)}+a}{4a}$$

$$x=\frac{-\sqrt{a\left(a-8\right)}+a}{4a}\text{, }a\neq 0$$

Solve for x

$$x=\frac{\sqrt{a\left(a-8\right)}+a}{4a}$$

$$x=\frac{-\sqrt{a\left(a-8\right)}+a}{4a}\text{, }a<0\text{ or }a\geq 8$$