Algebra Examples

$n^{2} + 7 n + 15 = 5$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Subtract $5$ from both sides of the equation.

$n^{2} + 7 n + 15 - 5 = 0$

Step 1.2

Subtract $5$ from $15$ .

$n^{2} + 7 n + 10 = 0$

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 $a = 1$ , $b = 7$ , and $c = 10$ into the quadratic formula and solve for $n$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $7$ to the power of $2$ .

$n = \frac{- 7 \pm \sqrt{49 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 10$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$n = \frac{- 7 \pm \sqrt{49 - 4 \cdot 10}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $10$ .

$n = \frac{- 7 \pm \sqrt{49 - 40}}{2 \cdot 1}$

Step 4.1.3

Subtract $40$ from $49$ .

$n = \frac{- 7 \pm \sqrt{9}}{2 \cdot 1}$

Step 4.1.4

Rewrite $9$ as $3^{2}$ .

$n = \frac{- 7 \pm \sqrt{3^{2}}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical, assuming positive real numbers.

$n = \frac{- 7 \pm 3}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$n = \frac{- 7 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$n = - 2, - 5$