Algebra Examples

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

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

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

Step 3.1.2

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

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Step 3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.1.2.2

Multiply $- 4$ by $- 3$ .

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

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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