Pre-Algebra Examples

$y = x^{2} - x - 13$

Step 1

Rewrite the equation as $x^{2} - x - 13 = y$ .

$x^{2} - x - 13 = y$

Step 2

Subtract $y$ from both sides of the equation.

$x^{2} - x - 13 - y = 0$

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 $a = 1$ , $b = - 1$ , and $c = - 13 - y$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot (- 13 - y))}}{2 \cdot 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 13 - y)}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot (- 13 - y)}}{2 \cdot 1}$

Step 5.1.3

Apply the distributive property.

$x = \frac{1 \pm \sqrt{1 - 4 \cdot - 13 - 4 (- y)}}{2 \cdot 1}$

Step 5.1.4

Multiply $- 4$ by $- 13$ .

$x = \frac{1 \pm \sqrt{1 + 52 - 4 (- y)}}{2 \cdot 1}$

Step 5.1.5

Multiply $- 1$ by $- 4$ .

$x = \frac{1 \pm \sqrt{1 + 52 + 4 y}}{2 \cdot 1}$

Step 5.1.6

Add $1$ and $52$ .

$x = \frac{1 \pm \sqrt{53 + 4 y}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{53 + 4 y}}{2}$

Step 6

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{53 + 4 y}}{2}$

$x = \frac{1 - \sqrt{53 + 4 y}}{2}$