Solve ax^2+bx-9=0

$$a x ^ { 2 } + b x - 9 = 0$$

Solve for a

$$a=-\frac{bx-9}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$ax^{2}-9=-bx$$

Add 9 to both sides.

$$ax^{2}=-bx+9$$

The equation is in standard form.

$$x^{2}a=9-bx$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{9-bx}{x^{2}}$$

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

$$a=\frac{9-bx}{x^{2}}$$

Solve for b

$$b=-ax+\frac{9}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$bx-9=-ax^{2}$$

Add 9 to both sides.

$$bx=-ax^{2}+9$$

The equation is in standard form.

$$xb=9-ax^{2}$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{9-ax^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{9-ax^{2}}{x}$$

Divide -ax^{2}+9 by x.

$$b=-ax+\frac{9}{x}$$