Basic Math Examples

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Multiply $- 4 \cdot 12 \cdot 3$ .

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Step 3.1.1.1

Multiply $- 4$ by $12$ .

$p = \frac{- b \pm \sqrt{b^{2} - 48 \cdot 3}}{2 \cdot 12}$

Step 3.1.1.2

Multiply $- 48$ by $3$ .

$p = \frac{- b \pm \sqrt{b^{2} - 144}}{2 \cdot 12}$

Step 3.1.2

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

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Step 3.1.2.1

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

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

Step 3.1.2.2

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 $b = 12$ .

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

Step 3.2

Multiply $2$ by $12$ .

$p = \frac{- b \pm \sqrt{(b + 12) (b - 12)}}{24}$

Step 4

The final answer is the combination of both solutions.

$p = - \frac{b - \sqrt{(b + 12) (b - 12)}}{24}$

$p = - \frac{b + \sqrt{(b + 12) (b - 12)}}{24}$