Precalculus Examples

$4 = 2 x^{2} + 4 x + 1$

Step 1

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

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Step 1.1

Move all the expressions to the left side of the equation.

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Step 1.1.1

Subtract $2 x^{2}$ from both sides of the equation.

$4 - 2 x^{2} = 4 x + 1$

Step 1.1.2

Subtract $4 x$ from both sides of the equation.

$4 - 2 x^{2} - 4 x = 1$

Step 1.1.3

Subtract $1$ from both sides of the equation.

$4 - 2 x^{2} - 4 x - 1 = 0$

Step 1.2

Subtract $1$ from $4$ .

$- 2 x^{2} - 4 x + 3 = 0$

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 $a = - 2$ , $b = - 4$ , and $c = 3$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 2 \cdot 3}}{2 \cdot - 2}$

Step 4.1.2

Multiply $- 4 \cdot - 2 \cdot 3$ .

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Step 4.1.2.1

Multiply $- 4$ by $- 2$ .

$x = \frac{4 \pm \sqrt{16 + 8 \cdot 3}}{2 \cdot - 2}$

Step 4.1.2.2

Multiply $8$ by $3$ .

$x = \frac{4 \pm \sqrt{16 + 24}}{2 \cdot - 2}$

Step 4.1.3

Add $16$ and $24$ .

$x = \frac{4 \pm \sqrt{40}}{2 \cdot - 2}$

Step 4.1.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 4.1.4.1

Factor $4$ out of $40$ .

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

Step 4.1.4.2

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot - 2}$

Step 4.1.5

Pull terms out from under the radical.

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

Step 4.2

Multiply $2$ by $- 2$ .

$x = \frac{4 \pm 2 \sqrt{10}}{- 4}$

Step 4.3

Simplify $\frac{4 \pm 2 \sqrt{10}}{- 4}$ .

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

Step 4.4

Move the negative in front of the fraction.

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 2.58113883 \dots, 0.58113883 \dots$