Solve 1-x-2x^2quadb=-2

$$1 - x - 2 x ^ { 2 } \quad b = - 2$$

Solve for b

$$b=\frac{3-x}{2x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply -1 and 2 to get -2.

$$1-x-2x^{2}b=-2$$

Subtract 1 from both sides.

$$-x-2x^{2}b=-2-1$$

Subtract 1 from -2 to get -3.

$$-x-2x^{2}b=-3$$

Add x to both sides.

$$-2x^{2}b=-3+x$$

The equation is in standard form.

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

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

$$\frac{\left(-2x^{2}\right)b}{-2x^{2}}=\frac{x-3}{-2x^{2}}$$

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

$$b=\frac{x-3}{-2x^{2}}$$

Divide x-3 by -2x^{2}.

$$b=-\frac{x-3}{2x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{24b+1}+1}{4b}\text{; }x=-\frac{-\sqrt{24b+1}+1}{4b}\text{, }&b\neq 0\\x=3\text{, }&b=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{24b+1}+1}{4b}\text{; }x=-\frac{-\sqrt{24b+1}+1}{4b}\text{, }&b\neq 0\text{ and }b\geq -\frac{1}{24}\\x=3\text{, }&b=0\end{matrix}\right.$$