Algebra Examples

x^{2} - 3 x - 4 = 0

$x^{2} - 3 x - 4 = 0$

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 = - 3

$b = - 3$ , and

c = - 4

$c = - 4$ into the quadratic formula and solve for

x

$x$ .

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

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

Step 3

Simplify.

Tap for more steps...

Step 3.1

Simplify the numerator.

Tap for more steps...

Step 3.1.1

Raise

- 3

$- 3$ to the power of

2

$2$ .

x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot - 4

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

Tap for more steps...

Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{3 \pm \sqrt{9 - 4 \cdot - 4}}{2 \cdot 1}

$x = \frac{3 \pm \sqrt{9 - 4 \cdot - 4}}{2 \cdot 1}$

Step 3.1.2.2

Multiply

- 4

$- 4$ by

- 4

$- 4$ .

x = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 1}

$x = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 1}$

x = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 1}

$x = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 1}$

Step 3.1.3

Add

9

$9$ and

16

$16$ .

x = \frac{3 \pm \sqrt{25}}{2 \cdot 1}

$x = \frac{3 \pm \sqrt{25}}{2 \cdot 1}$

Step 3.1.4

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

x = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 1}

$x = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 1}$

Step 3.1.5

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

x = \frac{3 \pm 5}{2 \cdot 1}

$x = \frac{3 \pm 5}{2 \cdot 1}$

x = \frac{3 \pm 5}{2 \cdot 1}

$x = \frac{3 \pm 5}{2 \cdot 1}$

Step 3.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{3 \pm 5}{2}

$x = \frac{3 \pm 5}{2}$

x = \frac{3 \pm 5}{2}

$x = \frac{3 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

x = 4, - 1

$x = 4, - 1$