Precalculus Examples

$x^{2} - 4 = 5 x$

Step 1

Subtract $5 x$ from both sides of the equation.

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

Step 2

Reorder terms.

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

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

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

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

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

Step 5.1.2

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

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot - 4}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 4$ .

$x = \frac{5 \pm \sqrt{25 + 16}}{2 \cdot 1}$

Step 5.1.3

Add $25$ and $16$ .

$x = \frac{5 \pm \sqrt{41}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$x = \frac{5 \pm \sqrt{41}}{2}$

Step 6

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

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

Step 6.1.2

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

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot - 4}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 4$ .

$x = \frac{5 \pm \sqrt{25 + 16}}{2 \cdot 1}$

Step 6.1.3

Add $25$ and $16$ .

$x = \frac{5 \pm \sqrt{41}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{5 \pm \sqrt{41}}{2}$

Step 6.3

Change the $\pm$ to $+$ .

$x = \frac{5 + \sqrt{41}}{2}$

Step 7

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

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

Step 7.1.2

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

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot - 4}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $- 4$ .

$x = \frac{5 \pm \sqrt{25 + 16}}{2 \cdot 1}$

Step 7.1.3

Add $25$ and $16$ .

$x = \frac{5 \pm \sqrt{41}}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$x = \frac{5 \pm \sqrt{41}}{2}$

Step 7.3

Change the $\pm$ to $-$ .

$x = \frac{5 - \sqrt{41}}{2}$

Step 8

The final answer is the combination of both solutions.

$x = \frac{5 + \sqrt{41}}{2}, \frac{5 - \sqrt{41}}{2}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5 + \sqrt{41}}{2}, \frac{5 - \sqrt{41}}{2}$

Decimal Form:

$x = 5.70156211 \dots, - 0.70156211 \dots$