Solve 2x^2+left(3-aright)x+1=0

$$2 { x }^{ 2 } + \left( 3-a \right) x+1 = 0$$

Solve for a

$$a=2x+3+\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3-a by x.

$$2x^{2}+3x-ax+1=0$$

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

$$3x-ax+1=-2x^{2}$$

Subtract 3x from both sides.

$$-ax+1=-2x^{2}-3x$$

Subtract 1 from both sides.

$$-ax=-2x^{2}-3x-1$$

The equation is in standard form.

$$\left(-x\right)a=-2x^{2}-3x-1$$

Divide both sides by -x.

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

Dividing by -x undoes the multiplication by -x.

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

Divide -\left(1+x\right)\left(1+2x\right) by -x.

$$a=2x+3+\frac{1}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{a^{2}-6a+1}+a-3}{4}$$

$$x=\frac{-\sqrt{a^{2}-6a+1}+a-3}{4}$$

Solve for x

$$x=\frac{\sqrt{a^{2}-6a+1}+a-3}{4}$$

$$x=\frac{-\sqrt{a^{2}-6a+1}+a-3}{4}\text{, }a\geq 2\sqrt{2}+3\text{ or }a\leq 3-2\sqrt{2}$$