Algebra Examples

2 x^{2} - 5 x - 3 = 0

$2 x^{2} - 5 x - 3 = 0$

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 = 2

$a = 2$ ,

b = - 5

$b = - 5$ , and

c = - 3

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

x

$x$ .

\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (2 \cdot - 3)}}{2 \cdot 2}

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (2 \cdot - 3)}}{2 \cdot 2}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

- 5

$- 5$ to the power of

2

$2$ .

x = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 3}}{2 \cdot 2}

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 3}}{2 \cdot 2}$

Step 3.1.2

Multiply

- 4 \cdot 2 \cdot - 3

$- 4 \cdot 2 \cdot - 3$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

2

$2$ .

x = \frac{5 \pm \sqrt{25 - 8 \cdot - 3}}{2 \cdot 2}

$x = \frac{5 \pm \sqrt{25 - 8 \cdot - 3}}{2 \cdot 2}$

Step 3.1.2.2

Multiply

- 8

$- 8$ by

- 3

$- 3$ .

x = \frac{5 \pm \sqrt{25 + 24}}{2 \cdot 2}

$x = \frac{5 \pm \sqrt{25 + 24}}{2 \cdot 2}$

x = \frac{5 \pm \sqrt{25 + 24}}{2 \cdot 2}

$x = \frac{5 \pm \sqrt{25 + 24}}{2 \cdot 2}$

Step 3.1.3

Add

25

$25$ and

24

$24$ .

x = \frac{5 \pm \sqrt{49}}{2 \cdot 2}

$x = \frac{5 \pm \sqrt{49}}{2 \cdot 2}$

Step 3.1.4

Rewrite

49

$49$ as

7^{2}

$7^{2}$ .

x = \frac{5 \pm \sqrt{7^{2}}}{2 \cdot 2}

$x = \frac{5 \pm \sqrt{7^{2}}}{2 \cdot 2}$

Step 3.1.5

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

x = \frac{5 \pm 7}{2 \cdot 2}

$x = \frac{5 \pm 7}{2 \cdot 2}$

x = \frac{5 \pm 7}{2 \cdot 2}

$x = \frac{5 \pm 7}{2 \cdot 2}$

Step 3.2

Multiply

2

$2$ by

2

$2$ .

x = \frac{5 \pm 7}{4}

$x = \frac{5 \pm 7}{4}$

x = \frac{5 \pm 7}{4}

$x = \frac{5 \pm 7}{4}$

Step 4

The final answer is the combination of both solutions.

x = 3, - \frac{1}{2}

$x = 3, - \frac{1}{2}$