Write as a Function of b 25p^2+bp+1=0

$25 p^{2} + b p + 1 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 25$ , $b = b$ , and $c = 1$ into the quadratic formula and solve for $p$ .

$\frac{- b \pm \sqrt{b^{2} - 4 \cdot (25 \cdot 1)}}{2 \cdot 25}$

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 25 \cdot 1$ as ${(2 \cdot 5 \cdot 1)}^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - {(2 \cdot 5 \cdot 1)}^{2}}}{2 \cdot 25}$

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 $b = 2 \cdot 5 \cdot 1$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 5 \cdot 1) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Simplify.

Multiply $2$ by $5$ .

$p = \frac{- b \pm \sqrt{(b + 10 \cdot 1) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Multiply $10$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Multiply $2$ by $5$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - (10 \cdot 1))}}{2 \cdot 25}$

Multiply $10$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 1 \cdot 10)}}{2 \cdot 25}$

Multiply $- 1$ by $10$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 10)}}{2 \cdot 25}$

Multiply $2$ by $25$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 10)}}{50}$

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

Simplify the numerator.

Rewrite $4 \cdot 25 \cdot 1$ as ${(2 \cdot 5 \cdot 1)}^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - {(2 \cdot 5 \cdot 1)}^{2}}}{2 \cdot 25}$

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 $b = 2 \cdot 5 \cdot 1$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 5 \cdot 1) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Simplify.

Multiply $2$ by $5$ .

$p = \frac{- b \pm \sqrt{(b + 10 \cdot 1) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Multiply $10$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Multiply $2$ by $5$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - (10 \cdot 1))}}{2 \cdot 25}$

Multiply $10$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 1 \cdot 10)}}{2 \cdot 25}$

Multiply $- 1$ by $10$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 10)}}{2 \cdot 25}$

Multiply $2$ by $25$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 10)}}{50}$

Change the $\pm$ to $+$ .

$p = \frac{- b + \sqrt{(b + 10) (b - 10)}}{50}$

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

$p = \frac{- (b) + \sqrt{(b + 10) (b - 10)}}{50}$

Factor $- 1$ out of $\sqrt{(b + 10) (b - 10)}$ .

$p = \frac{- (b) - 1 (- \sqrt{(b + 10) (b - 10)})}{50}$

Factor $- 1$ out of $- (b) - 1 (- \sqrt{(b + 10) (b - 10)})$ .

$p = \frac{- (b - \sqrt{(b + 10) (b - 10)})}{50}$

Rewrite $- (b - \sqrt{(b + 10) (b - 10)})$ as $- 1 (b - \sqrt{(b + 10) (b - 10)})$ .

$p = \frac{- 1 (b - \sqrt{(b + 10) (b - 10)})}{50}$

Move the negative in front of the fraction.

$p = - \frac{b - \sqrt{(b + 10) (b - 10)}}{50}$

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

Simplify the numerator.

Rewrite $4 \cdot 25 \cdot 1$ as ${(2 \cdot 5 \cdot 1)}^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - {(2 \cdot 5 \cdot 1)}^{2}}}{2 \cdot 25}$

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 $b = 2 \cdot 5 \cdot 1$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 5 \cdot 1) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Simplify.

Multiply $2$ by $5$ .

$p = \frac{- b \pm \sqrt{(b + 10 \cdot 1) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Multiply $10$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - (2 \cdot 5 \cdot 1))}}{2 \cdot 25}$

Multiply $2$ by $5$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - (10 \cdot 1))}}{2 \cdot 25}$

Multiply $10$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 1 \cdot 10)}}{2 \cdot 25}$

Multiply $- 1$ by $10$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 10)}}{2 \cdot 25}$

Multiply $2$ by $25$ .

$p = \frac{- b \pm \sqrt{(b + 10) (b - 10)}}{50}$

Change the $\pm$ to $-$ .

$p = \frac{- b - \sqrt{(b + 10) (b - 10)}}{50}$

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

$p = \frac{- (b) - \sqrt{(b + 10) (b - 10)}}{50}$

Factor $- 1$ out of $- \sqrt{(b + 10) (b - 10)}$ .

$p = \frac{- (b) - (\sqrt{(b + 10) (b - 10)})}{50}$

Factor $- 1$ out of $- (b) - (\sqrt{(b + 10) (b - 10)})$ .

$p = \frac{- (b + \sqrt{(b + 10) (b - 10)})}{50}$

Rewrite $- (b + \sqrt{(b + 10) (b - 10)})$ as $- 1 (b + \sqrt{(b + 10) (b - 10)})$ .

$p = \frac{- 1 (b + \sqrt{(b + 10) (b - 10)})}{50}$

Move the negative in front of the fraction.

$p = - \frac{b + \sqrt{(b + 10) (b - 10)}}{50}$

The final answer is the combination of both solutions.

$p = - \frac{b - \sqrt{(b + 10) (b - 10)}}{50}$

$p = - \frac{b + \sqrt{(b + 10) (b - 10)}}{50}$

Write as a Function of b 25p^2+bp+1=0

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