Solve (a-1)x^2+(b+1)x+2+c=0

$$( a - 1 ) x ^ { 2 } + ( b + 1 ) x + 2 + c = 0$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{2+c+x+bx-x^{2}}{x^{2}}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&c=-2\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-1 by x^{2}.

$$ax^{2}-x^{2}+\left(b+1\right)x+2+c=0$$

Use the distributive property to multiply b+1 by x.

$$ax^{2}-x^{2}+bx+x+2+c=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$ax^{2}+bx+x+2+c=x^{2}$$

Subtract bx from both sides.

$$ax^{2}+x+2+c=x^{2}-bx$$

Subtract x from both sides.

$$ax^{2}+2+c=x^{2}-bx-x$$

Subtract 2 from both sides.

$$ax^{2}+c=x^{2}-bx-x-2$$

Subtract c from both sides.

$$ax^{2}=x^{2}-bx-x-2-c$$

The equation is in standard form.

$$x^{2}a=x^{2}-bx-x-c-2$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{x^{2}-bx-x-c-2}{x^{2}}$$

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

$$a=\frac{x^{2}-bx-x-c-2}{x^{2}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-ax+x-\frac{c}{x}-1-\frac{2}{x}\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&c=-2\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-1 by x^{2}.

$$ax^{2}-x^{2}+\left(b+1\right)x+2+c=0$$

Use the distributive property to multiply b+1 by x.

$$ax^{2}-x^{2}+bx+x+2+c=0$$

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

$$-x^{2}+bx+x+2+c=-ax^{2}$$

Add x^{2} to both sides.

$$bx+x+2+c=-ax^{2}+x^{2}$$

Subtract x from both sides.

$$bx+2+c=-ax^{2}+x^{2}-x$$

Subtract 2 from both sides.

$$bx+c=-ax^{2}+x^{2}-x-2$$

Subtract c from both sides.

$$bx=-ax^{2}+x^{2}-x-2-c$$

Reorder the terms.

$$bx=-ax^{2}+x^{2}-x-c-2$$

The equation is in standard form.

$$xb=-ax^{2}+x^{2}-x-c-2$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-ax^{2}+x^{2}-x-c-2}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-ax^{2}+x^{2}-x-c-2}{x}$$

Divide -ax^{2}+x^{2}-x-c-2 by x.

$$b=-ax+x-\frac{c+2}{x}-1$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{2+c+x+bx-x^{2}}{x^{2}}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&c=-2\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-1 by x^{2}.

$$ax^{2}-x^{2}+\left(b+1\right)x+2+c=0$$

Use the distributive property to multiply b+1 by x.

$$ax^{2}-x^{2}+bx+x+2+c=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$ax^{2}+bx+x+2+c=x^{2}$$

Subtract bx from both sides.

$$ax^{2}+x+2+c=x^{2}-bx$$

Subtract x from both sides.

$$ax^{2}+2+c=x^{2}-bx-x$$

Subtract 2 from both sides.

$$ax^{2}+c=x^{2}-bx-x-2$$

Subtract c from both sides.

$$ax^{2}=x^{2}-bx-x-2-c$$

The equation is in standard form.

$$x^{2}a=x^{2}-bx-x-c-2$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{x^{2}-bx-x-c-2}{x^{2}}$$

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

$$a=\frac{x^{2}-bx-x-c-2}{x^{2}}$$

Solve for b

$$\left\{\begin{matrix}b=-ax+x-\frac{c}{x}-1-\frac{2}{x}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&c=-2\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-1 by x^{2}.

$$ax^{2}-x^{2}+\left(b+1\right)x+2+c=0$$

Use the distributive property to multiply b+1 by x.

$$ax^{2}-x^{2}+bx+x+2+c=0$$

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

$$-x^{2}+bx+x+2+c=-ax^{2}$$

Add x^{2} to both sides.

$$bx+x+2+c=-ax^{2}+x^{2}$$

Subtract x from both sides.

$$bx+2+c=-ax^{2}+x^{2}-x$$

Subtract 2 from both sides.

$$bx+c=-ax^{2}+x^{2}-x-2$$

Subtract c from both sides.

$$bx=-ax^{2}+x^{2}-x-2-c$$

Reorder the terms.

$$bx=-ax^{2}+x^{2}-x-c-2$$

The equation is in standard form.

$$xb=-ax^{2}+x^{2}-x-c-2$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-ax^{2}+x^{2}-x-c-2}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-ax^{2}+x^{2}-x-c-2}{x}$$

Divide -ax^{2}+x^{2}-x-c-2 by x.

$$b=-ax+x-\frac{c+2}{x}-1$$