Basic Math Examples

$b^{2} + b + c + 3 = 2 b$

Step 1

Move all terms containing $b$ to the left side of the equation.

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Step 1.1

Subtract $2 b$ from both sides of the equation.

$b^{2} + b + c + 3 - 2 b = 0$

Step 1.2

Subtract $2 b$ from $b$ .

$b^{2} - b + c + 3 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 1$ , and $c = c + 3$ into the quadratic formula and solve for $b$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot (c + 3))}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 1$ to the power of $2$ .

$b = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot (c + 3)}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4$ by $1$ .

$b = \frac{1 \pm \sqrt{1 - 4 \cdot (c + 3)}}{2 \cdot 1}$

Step 4.1.3

Apply the distributive property.

$b = \frac{1 \pm \sqrt{1 - 4 c - 4 \cdot 3}}{2 \cdot 1}$

Step 4.1.4

Multiply $- 4$ by $3$ .

$b = \frac{1 \pm \sqrt{1 - 4 c - 12}}{2 \cdot 1}$

Step 4.1.5

Subtract $12$ from $1$ .

$b = \frac{1 \pm \sqrt{- 4 c - 11}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$b = \frac{1 \pm \sqrt{- 4 c - 11}}{2}$

Step 5

The final answer is the combination of both solutions.

$b = \frac{1 + \sqrt{- 4 c - 11}}{2}$

$b = \frac{1 - \sqrt{- 4 c - 11}}{2}$