Precalculus Examples

$2 x^{2} - (x + 2) (x - 3) = 12$

Step 1

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

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Step 1.1

Simplify the left side.

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Step 1.1.1

Simplify $2 x^{2} - (x + 2) (x - 3)$ .

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Step 1.1.1.1

Simplify each term.

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Step 1.1.1.1.1

Apply the distributive property.

$2 x^{2} + (- x - 1 \cdot 2) (x - 3) = 12$

Step 1.1.1.1.2

Multiply $- 1$ by $2$ .

$2 x^{2} + (- x - 2) (x - 3) = 12$

Step 1.1.1.1.3

Expand $(- x - 2) (x - 3)$ using the FOIL Method.

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Step 1.1.1.1.3.1

Apply the distributive property.

$2 x^{2} - x (x - 3) - 2 (x - 3) = 12$

Step 1.1.1.1.3.2

Apply the distributive property.

$2 x^{2} - x \cdot x - x \cdot - 3 - 2 (x - 3) = 12$

Step 1.1.1.1.3.3

Apply the distributive property.

$2 x^{2} - x \cdot x - x \cdot - 3 - 2 x - 2 \cdot - 3 = 12$

Step 1.1.1.1.4

Simplify and combine like terms.

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Step 1.1.1.1.4.1

Simplify each term.

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Step 1.1.1.1.4.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.1.1.4.1.1.1

Move $x$ .

$2 x^{2} - (x \cdot x) - x \cdot - 3 - 2 x - 2 \cdot - 3 = 12$

Step 1.1.1.1.4.1.1.2

Multiply $x$ by $x$ .

$2 x^{2} - x^{2} - x \cdot - 3 - 2 x - 2 \cdot - 3 = 12$

Step 1.1.1.1.4.1.2

Multiply $- 3$ by $- 1$ .

$2 x^{2} - x^{2} + 3 x - 2 x - 2 \cdot - 3 = 12$

Step 1.1.1.1.4.1.3

Multiply $- 2$ by $- 3$ .

$2 x^{2} - x^{2} + 3 x - 2 x + 6 = 12$

Step 1.1.1.1.4.2

Subtract $2 x$ from $3 x$ .

$2 x^{2} - x^{2} + x + 6 = 12$

Step 1.1.1.2

Subtract $x^{2}$ from $2 x^{2}$ .

$x^{2} + x + 6 = 12$

Step 1.2

Subtract $12$ from both sides of the equation.

$x^{2} + x + 6 - 12 = 0$

Step 1.3

Subtract $12$ from $6$ .

$x^{2} + x - 6 = 0$

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 $a = 1$ , $b = 1$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 6)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 6}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 1 \pm \sqrt{1 + 24}}{2 \cdot 1}$

Step 4.1.3

Add $1$ and $24$ .

$x = \frac{- 1 \pm \sqrt{25}}{2 \cdot 1}$

Step 4.1.4

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 1 \pm \sqrt{5^{2}}}{2 \cdot 1}$

Step 4.1.5

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

$x = \frac{- 1 \pm 5}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 5}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 2, - 3$