Algebra Examples

$\frac{- 6 + - \sqrt{6^{2} - 4 \cdot 3 \cdot - 9}}{2 (3)}$

Step 1

Replace all occurrences of $+$ $-$ with a single $-$ . A plus sign followed by a minus sign has the same mathematical meaning as a single minus sign because $1 \cdot - 1 = - 1$

$\frac{- 6 - \sqrt{6^{2} - 4 \cdot 3 \cdot - 9}}{2 (3)}$

Step 2

Set $\frac{- 6 + - \sqrt{6^{2} - 4 \cdot 3 \cdot - 9}}{2 (3)}$ equal to $0$ .

$\frac{- 6 + - \sqrt{6^{2} - 4 \cdot 3 \cdot - 9}}{2 (3)} = 0$

Step 3

Simplify the left side.

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Step 3.1

Simplify $\frac{- 6 + - \sqrt{6^{2} - 4 \cdot 3 \cdot - 9}}{2 (3)}$ .

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Step 3.1.1

Simplify the numerator.

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Step 3.1.1.1

Raise $6$ to the power of $2$ .

$\frac{- 6 - \sqrt{36 - 4 \cdot 3 \cdot - 9}}{2 (3)} = 0$

Step 3.1.1.2

Multiply $- 4 \cdot 3 \cdot - 9$ .

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Step 3.1.1.2.1

Multiply $- 4$ by $3$ .

$\frac{- 6 - \sqrt{36 - 12 \cdot - 9}}{2 (3)} = 0$

Step 3.1.1.2.2

Multiply $- 12$ by $- 9$ .

$\frac{- 6 - \sqrt{36 + 108}}{2 (3)} = 0$

Step 3.1.1.3

Add $36$ and $108$ .

$\frac{- 6 - \sqrt{144}}{2 (3)} = 0$

Step 3.1.1.4

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

$\frac{- 6 - \sqrt{12^{2}}}{2 (3)} = 0$

Step 3.1.1.5

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

$\frac{- 6 - 1 \cdot 12}{2 (3)} = 0$

Step 3.1.1.6

Multiply $- 1$ by $12$ .

$\frac{- 6 - 12}{2 (3)} = 0$

Step 3.1.1.7

Subtract $12$ from $- 6$ .

$\frac{- 18}{2 (3)} = 0$

Step 3.1.2

Simplify the expression.

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Step 3.1.2.1

Multiply $2$ by $3$ .

$\frac{- 18}{6} = 0$

Step 3.1.2.2

Divide $- 18$ by $6$ .

$- 3 = 0$

Step 4

Since $- 3 \neq 0$ , there are no solutions.

No solution