Write as a Function of b 3p^2+bp+12=0

$3 p^{2} + b p + 12 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = b$ , and $c = 12$ into the quadratic formula and solve for $p$ .

$\frac{- b \pm \sqrt{b^{2} - 4 \cdot (3 \cdot 12)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Multiply $- 4$ by $3$ .

$p = \frac{- b \pm \sqrt{b^{2} - 12 \cdot 12}}{2 \cdot 3}$

Multiply $- 12$ by $12$ .

$p = \frac{- b \pm \sqrt{b^{2} - 144}}{2 \cdot 3}$

Rewrite $b^{2} - 144$ in a factored form.

Rewrite $144$ as $12^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - 12^{2}}}{2 \cdot 3}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 12$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{6}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Multiply $- 4$ by $3$ .

$p = \frac{- b \pm \sqrt{b^{2} - 12 \cdot 12}}{2 \cdot 3}$

Multiply $- 12$ by $12$ .

$p = \frac{- b \pm \sqrt{b^{2} - 144}}{2 \cdot 3}$

Rewrite $b^{2} - 144$ in a factored form.

Rewrite $144$ as $12^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - 12^{2}}}{2 \cdot 3}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 12$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{6}$

Change the $\pm$ to $+$ .

$p = \frac{- b + \sqrt{(b + 12) (b - 12)}}{6}$

Factor $- 1$ out of $- b$ .

$p = \frac{- (b) + \sqrt{(b + 12) (b - 12)}}{6}$

Factor $- 1$ out of $\sqrt{(b + 12) (b - 12)}$ .

$p = \frac{- (b) - 1 (- \sqrt{(b + 12) (b - 12)})}{6}$

Factor $- 1$ out of $- (b) - 1 (- \sqrt{(b + 12) (b - 12)})$ .

$p = \frac{- (b - \sqrt{(b + 12) (b - 12)})}{6}$

Rewrite $- (b - \sqrt{(b + 12) (b - 12)})$ as $- 1 (b - \sqrt{(b + 12) (b - 12)})$ .

$p = \frac{- 1 (b - \sqrt{(b + 12) (b - 12)})}{6}$

Move the negative in front of the fraction.

$p = - \frac{b - \sqrt{(b + 12) (b - 12)}}{6}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Multiply $- 4$ by $3$ .

$p = \frac{- b \pm \sqrt{b^{2} - 12 \cdot 12}}{2 \cdot 3}$

Multiply $- 12$ by $12$ .

$p = \frac{- b \pm \sqrt{b^{2} - 144}}{2 \cdot 3}$

Rewrite $b^{2} - 144$ in a factored form.

Rewrite $144$ as $12^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - 12^{2}}}{2 \cdot 3}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 12$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{6}$

Change the $\pm$ to $-$ .

$p = \frac{- b - \sqrt{(b + 12) (b - 12)}}{6}$

Factor $- 1$ out of $- b$ .

$p = \frac{- (b) - \sqrt{(b + 12) (b - 12)}}{6}$

Factor $- 1$ out of $- \sqrt{(b + 12) (b - 12)}$ .

$p = \frac{- (b) - (\sqrt{(b + 12) (b - 12)})}{6}$

Factor $- 1$ out of $- (b) - (\sqrt{(b + 12) (b - 12)})$ .

$p = \frac{- (b + \sqrt{(b + 12) (b - 12)})}{6}$

Rewrite $- (b + \sqrt{(b + 12) (b - 12)})$ as $- 1 (b + \sqrt{(b + 12) (b - 12)})$ .

$p = \frac{- 1 (b + \sqrt{(b + 12) (b - 12)})}{6}$

Move the negative in front of the fraction.

$p = - \frac{b + \sqrt{(b + 12) (b - 12)}}{6}$

The final answer is the combination of both solutions.

$p = - \frac{b - \sqrt{(b + 12) (b - 12)}}{6}$

$p = - \frac{b + \sqrt{(b + 12) (b - 12)}}{6}$

Write as a Function of b 3p^2+bp+12=0

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