Pre-Algebra Examples

$y = x^{2} + 2 x - 3$

Step 1

Rewrite the equation as $x^{2} + 2 x - 3 = y$ .

$x^{2} + 2 x - 3 = y$

Step 2

Subtract $y$ from both sides of the equation.

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

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 $a = 1$ , $b = 2$ , and $c = - 3 - y$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot (- 3 - y))}}{2 \cdot 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 3 - y)}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot (- 3 - y)}}{2 \cdot 1}$

Step 5.1.3

Apply the distributive property.

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 3 - 4 (- y)}}{2 \cdot 1}$

Step 5.1.4

Multiply $- 4$ by $- 3$ .

$x = \frac{- 2 \pm \sqrt{4 + 12 - 4 (- y)}}{2 \cdot 1}$

Step 5.1.5

Multiply $- 1$ by $- 4$ .

$x = \frac{- 2 \pm \sqrt{4 + 12 + 4 y}}{2 \cdot 1}$

Step 5.1.6

Add $4$ and $12$ .

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

Step 5.1.7

Factor $4$ out of $16 + 4 y$ .

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Step 5.1.7.1

Factor $4$ out of $16$ .

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

Step 5.1.7.2

Factor $4$ out of $4 \cdot 4 + 4 y$ .

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

Step 5.1.8

Rewrite $4 (4 + y)$ as $2^{2} (2^{2} + y)$ .

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Step 5.1.8.1

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

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

Step 5.1.8.2

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

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

Step 5.1.9

Pull terms out from under the radical.

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

Step 5.1.10

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

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

Step 5.2

Multiply $2$ by $1$ .

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

Step 5.3

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

$x = - 1 \pm \sqrt{4 + y}$

Step 6

The final answer is the combination of both solutions.

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

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