Pre-Algebra Examples

$4 = - 20 x^{2} + 16 x$

Step 1

Rewrite the equation as $- 20 x^{2} + 16 x = 4$ .

$- 20 x^{2} + 16 x = 4$

Step 2

Subtract $4$ from both sides of the equation.

$- 20 x^{2} + 16 x - 4 = 0$

Step 3

Factor $- 4$ out of $- 20 x^{2} + 16 x - 4$ .

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Step 3.1

Factor $- 4$ out of $- 20 x^{2}$ .

$- 4 (5 x^{2}) + 16 x - 4 = 0$

Step 3.2

Factor $- 4$ out of $16 x$ .

$- 4 (5 x^{2}) - 4 (- 4 x) - 4 = 0$

Step 3.3

Factor $- 4$ out of $- 4$ .

$- 4 (5 x^{2}) - 4 (- 4 x) - 4 \cdot 1 = 0$

Step 3.4

Factor $- 4$ out of $- 4 (5 x^{2}) - 4 (- 4 x)$ .

$- 4 (5 x^{2} - 4 x) - 4 \cdot 1 = 0$

Step 3.5

Factor $- 4$ out of $- 4 (5 x^{2} - 4 x) - 4 (1)$ .

$- 4 (5 x^{2} - 4 x + 1) = 0$

Step 4

Divide each term in $- 4 (5 x^{2} - 4 x + 1) = 0$ by $- 4$ and simplify.

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Step 4.1

Divide each term in $- 4 (5 x^{2} - 4 x + 1) = 0$ by $- 4$ .

$\frac{- 4 (5 x^{2} - 4 x + 1)}{- 4} = \frac{0}{- 4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 4 (5 x^{2} - 4 x + 1)}{- 4} = \frac{0}{- 4}$

Step 4.2.1.2

Divide $5 x^{2} - 4 x + 1$ by $1$ .

$5 x^{2} - 4 x + 1 = \frac{0}{- 4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 4$ .

$5 x^{2} - 4 x + 1 = 0$

Step 5

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6

Substitute the values $a = 5$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (5 \cdot 1)}}{2 \cdot 5}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 5 \cdot 1}}{2 \cdot 5}$

Step 7.1.2

Multiply $- 4 \cdot 5 \cdot 1$ .

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Step 7.1.2.1

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20 \cdot 1}}{2 \cdot 5}$

Step 7.1.2.2

Multiply $- 20$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 20}}{2 \cdot 5}$

Step 7.1.3

Subtract $20$ from $16$ .

$x = \frac{4 \pm \sqrt{- 4}}{2 \cdot 5}$

Step 7.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 5}$

Step 7.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 5}$

Step 7.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{4}}{2 \cdot 5}$

Step 7.1.7

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 5}$

Step 7.1.8

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

$x = \frac{4 \pm i \cdot 2}{2 \cdot 5}$

Step 7.1.9

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i}{2 \cdot 5}$

Step 7.2

Multiply $2$ by $5$ .

$x = \frac{4 \pm 2 i}{10}$

Step 7.3

Simplify $\frac{4 \pm 2 i}{10}$ .

$x = \frac{2 \pm i}{5}$

Step 8

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 8.1

Simplify the numerator.

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Step 8.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 5 \cdot 1}}{2 \cdot 5}$

Step 8.1.2

Multiply $- 4 \cdot 5 \cdot 1$ .

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Step 8.1.2.1

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20 \cdot 1}}{2 \cdot 5}$

Step 8.1.2.2

Multiply $- 20$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 20}}{2 \cdot 5}$

Step 8.1.3

Subtract $20$ from $16$ .

$x = \frac{4 \pm \sqrt{- 4}}{2 \cdot 5}$

Step 8.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 5}$

Step 8.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 5}$

Step 8.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{4}}{2 \cdot 5}$

Step 8.1.7

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 5}$

Step 8.1.8

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

$x = \frac{4 \pm i \cdot 2}{2 \cdot 5}$

Step 8.1.9

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i}{2 \cdot 5}$

Step 8.2

Multiply $2$ by $5$ .

$x = \frac{4 \pm 2 i}{10}$

Step 8.3

Simplify $\frac{4 \pm 2 i}{10}$ .

$x = \frac{2 \pm i}{5}$

Step 8.4

Change the $\pm$ to $+$ .

$x = \frac{2 + i}{5}$

Step 8.5

Split the fraction $\frac{2 + i}{5}$ into two fractions.

$x = \frac{2}{5} + \frac{i}{5}$

Step 9

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 9.1

Simplify the numerator.

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Step 9.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 5 \cdot 1}}{2 \cdot 5}$

Step 9.1.2

Multiply $- 4 \cdot 5 \cdot 1$ .

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Step 9.1.2.1

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20 \cdot 1}}{2 \cdot 5}$

Step 9.1.2.2

Multiply $- 20$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 20}}{2 \cdot 5}$

Step 9.1.3

Subtract $20$ from $16$ .

$x = \frac{4 \pm \sqrt{- 4}}{2 \cdot 5}$

Step 9.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 5}$

Step 9.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 5}$

Step 9.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{4}}{2 \cdot 5}$

Step 9.1.7

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 5}$

Step 9.1.8

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

$x = \frac{4 \pm i \cdot 2}{2 \cdot 5}$

Step 9.1.9

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i}{2 \cdot 5}$

Step 9.2

Multiply $2$ by $5$ .

$x = \frac{4 \pm 2 i}{10}$

Step 9.3

Simplify $\frac{4 \pm 2 i}{10}$ .

$x = \frac{2 \pm i}{5}$

Step 9.4

Change the $\pm$ to $-$ .

$x = \frac{2 - i}{5}$

Step 9.5

Split the fraction $\frac{2 - i}{5}$ into two fractions.

$x = \frac{2}{5} + \frac{- i}{5}$

Step 9.6

Move the negative in front of the fraction.

$x = \frac{2}{5} - \frac{i}{5}$

Step 10

The final answer is the combination of both solutions.

$x = \frac{2}{5} + \frac{i}{5}, \frac{2}{5} - \frac{i}{5}$