Chemistry Examples

$62 = x (x + 10)$

Step 1

Rewrite the equation as $x (x + 10) = 62$ .

$x (x + 10) = 62$

Step 2

Simplify $x (x + 10)$ .

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Step 2.1

Apply the distributive property.

$x \cdot x + x \cdot 10 = 62$

Step 2.2

Simplify the expression.

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Step 2.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot 10 = 62$

Step 2.2.2

Move $10$ to the left of $x$ .

$x^{2} + 10 x = 62$

Step 3

Subtract $62$ from both sides of the equation.

$x^{2} + 10 x - 62 = 0$

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 $a = 1$ , $b = 10$ , and $c = - 62$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot - 62)}}{2 \cdot 1}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot - 62}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot - 62$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot - 62}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 62$ .

$x = \frac{- 10 \pm \sqrt{100 + 248}}{2 \cdot 1}$

Step 6.1.3

Add $100$ and $248$ .

$x = \frac{- 10 \pm \sqrt{348}}{2 \cdot 1}$

Step 6.1.4

Rewrite $348$ as $2^{2} \cdot 87$ .

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Step 6.1.4.1

Factor $4$ out of $348$ .

$x = \frac{- 10 \pm \sqrt{4 (87)}}{2 \cdot 1}$

Step 6.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 10 \pm \sqrt{2^{2} \cdot 87}}{2 \cdot 1}$

Step 6.1.5

Pull terms out from under the radical.

$x = \frac{- 10 \pm 2 \sqrt{87}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{- 10 \pm 2 \sqrt{87}}{2}$

Step 6.3

Simplify $\frac{- 10 \pm 2 \sqrt{87}}{2}$ .

$x = - 5 \pm \sqrt{87}$

Step 7

The final answer is the combination of both solutions.

$x = - 5 + \sqrt{87}, - 5 - \sqrt{87}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = - 5 + \sqrt{87}, - 5 - \sqrt{87}$

Decimal Form:

$x = 4.32737905 \dots, - 14.32737905 \dots$