Solve for b x^2+bx+c=0

$x^{2} + b x + c = 0$

Step 1

Move all terms not containing $b$ to the right side of the equation.

Subtract $x^{2}$ from both sides of the equation.

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

Subtract $c$ from both sides of the equation.

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

Step 2

Divide each term in $b x = - x^{2} - c$ by $x$ and simplify.

Divide each term in $b x = - x^{2} - c$ by $x$ .

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

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $x^{2}$ and $x$ .

Factor $x$ out of $- x^{2}$ .

$b = \frac{x (- x)}{x} + \frac{- c}{x}$

Cancel the common factors.

Raise $x$ to the power of $1$ .

$b = \frac{x (- x)}{x^{1}} + \frac{- c}{x}$

Factor $x$ out of $x^{1}$ .

$b = \frac{x (- x)}{x \cdot 1} + \frac{- c}{x}$

Cancel the common factor.

$b = \frac{x (- x)}{x \cdot 1} + \frac{- c}{x}$

Rewrite the expression.

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

Divide $- x$ by $1$ .

$b = - x + \frac{- c}{x}$

Move the negative in front of the fraction.

$b = - x - \frac{c}{x}$