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

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

Step 1

Rewrite the equation as $2 x^{2} + b x + c = y$ .

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

Step 2

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

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

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

Subtract $c$ from both sides of the equation.

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

Step 3

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

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

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

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

Simplify the right side.

Simplify each term.

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

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

$b = \frac{y}{x} + \frac{x (- 2 x)}{x} + \frac{- c}{x}$

Cancel the common factors.

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

$b = \frac{y}{x} + \frac{x (- 2 x)}{x^{1}} + \frac{- c}{x}$

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

$b = \frac{y}{x} + \frac{x (- 2 x)}{x \cdot 1} + \frac{- c}{x}$

Cancel the common factor.

$b = \frac{y}{x} + \frac{x (- 2 x)}{x \cdot 1} + \frac{- c}{x}$

Rewrite the expression.

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

Divide $- 2 x$ by $1$ .

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

Move the negative in front of the fraction.

$b = \frac{y}{x} - 2 x - \frac{c}{x}$