Finite Math Examples

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

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

Step 1

Use the quadratic formula to find the solutions.

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

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

Step 2

Substitute the values

a = 1

$a = 1$ ,

b = 2

$b = 2$ , and

c = - 3

$c = - 3$ into the quadratic formula and solve for

x

$x$ .

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

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

2

$2$ to the power of

2

$2$ .

x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}

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

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot - 3

$- 4 \cdot 1 \cdot - 3$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 3}}{2 \cdot 1}

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

Step 3.1.2.2

Multiply

- 4

$- 4$ by

- 3

$- 3$ .

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

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

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

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

Step 3.1.3

Add

$4$ and

$12$ .

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

Step 3.1.4

Rewrite

$16$ as

$4^{2}$ .

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

Step 3.1.5

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

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

Step 3.2

Multiply

$2$ by

$1$ .

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

Step 3.3

Simplify

$\frac{- 2 \pm 4}{2}$ .

$x = - 1 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 3$

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