Basic Math Examples

$p = 5 (5 - p) (3 p + 1)$

Step 1

Since $p$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$5 (5 - p) (3 p + 1) = p$

Step 2

Simplify $5 (5 - p) (3 p + 1)$ .

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Step 2.1

Rewrite.

$0 + 0 + 5 (5 - p) (3 p + 1) = p$

Step 2.2

Simplify by multiplying through.

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Step 2.2.1

Apply the distributive property.

$(5 \cdot 5 + 5 (- p)) (3 p + 1) = p$

Step 2.2.2

Multiply.

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Step 2.2.2.1

Multiply $5$ by $5$ .

$(25 + 5 (- p)) (3 p + 1) = p$

Step 2.2.2.2

Multiply $- 1$ by $5$ .

$(25 - 5 p) (3 p + 1) = p$

Step 2.3

Expand $(25 - 5 p) (3 p + 1)$ using the FOIL Method.

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Step 2.3.1

Apply the distributive property.

$25 (3 p + 1) - 5 p (3 p + 1) = p$

Step 2.3.2

Apply the distributive property.

$25 (3 p) + 25 \cdot 1 - 5 p (3 p + 1) = p$

Step 2.3.3

Apply the distributive property.

$25 (3 p) + 25 \cdot 1 - 5 p (3 p) - 5 p \cdot 1 = p$

Step 2.4

Simplify and combine like terms.

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Step 2.4.1

Simplify each term.

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Step 2.4.1.1

Multiply $3$ by $25$ .

$75 p + 25 \cdot 1 - 5 p (3 p) - 5 p \cdot 1 = p$

Step 2.4.1.2

Multiply $25$ by $1$ .

$75 p + 25 - 5 p (3 p) - 5 p \cdot 1 = p$

Step 2.4.1.3

Rewrite using the commutative property of multiplication.

$75 p + 25 - 5 \cdot 3 p \cdot p - 5 p \cdot 1 = p$

Step 2.4.1.4

Multiply $p$ by $p$ by adding the exponents.

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Step 2.4.1.4.1

Move $p$ .

$75 p + 25 - 5 \cdot 3 (p \cdot p) - 5 p \cdot 1 = p$

Step 2.4.1.4.2

Multiply $p$ by $p$ .

$75 p + 25 - 5 \cdot 3 p^{2} - 5 p \cdot 1 = p$

Step 2.4.1.5

Multiply $- 5$ by $3$ .

$75 p + 25 - 15 p^{2} - 5 p \cdot 1 = p$

Step 2.4.1.6

Multiply $- 5$ by $1$ .

$75 p + 25 - 15 p^{2} - 5 p = p$

Step 2.4.2

Subtract $5 p$ from $75 p$ .

$70 p + 25 - 15 p^{2} = p$

Step 3

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

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Step 3.1

Subtract $p$ from both sides of the equation.

$70 p + 25 - 15 p^{2} - p = 0$

Step 3.2

Subtract $p$ from $70 p$ .

$69 p + 25 - 15 p^{2} = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = - 15$ , $b = 69$ , and $c = 25$ into the quadratic formula and solve for $p$ .

$\frac{- 69 \pm \sqrt{69^{2} - 4 \cdot (- 15 \cdot 25)}}{2 \cdot - 15}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $69$ to the power of $2$ .

$p = \frac{- 69 \pm \sqrt{4761 - 4 \cdot - 15 \cdot 25}}{2 \cdot - 15}$

Step 6.1.2

Multiply $- 4 \cdot - 15 \cdot 25$ .

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Step 6.1.2.1

Multiply $- 4$ by $- 15$ .

$p = \frac{- 69 \pm \sqrt{4761 + 60 \cdot 25}}{2 \cdot - 15}$

Step 6.1.2.2

Multiply $60$ by $25$ .

$p = \frac{- 69 \pm \sqrt{4761 + 1500}}{2 \cdot - 15}$

Step 6.1.3

Add $4761$ and $1500$ .

$p = \frac{- 69 \pm \sqrt{6261}}{2 \cdot - 15}$

Step 6.2

Multiply $2$ by $- 15$ .

$p = \frac{- 69 \pm \sqrt{6261}}{- 30}$

Step 6.3

Simplify $\frac{- 69 \pm \sqrt{6261}}{- 30}$ .

$p = \frac{69 \pm \sqrt{6261}}{30}$

Step 7

The final answer is the combination of both solutions.

$p = \frac{69 + \sqrt{6261}}{30}, \frac{69 - \sqrt{6261}}{30}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$p = \frac{69 + \sqrt{6261}}{30}, \frac{69 - \sqrt{6261}}{30}$

Decimal Form:

$p = 4.93754936 \dots, - 0.33754936 \dots$