Algebra Examples

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

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

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 = - 2

$c = - 2$ into the quadratic formula and solve for

r

$r$ .

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

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

3

$3$ to the power of

2

$2$ .

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

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

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot - 2

$- 4 \cdot 1 \cdot - 2$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 3.1.2.2

Multiply

- 4

$- 4$ by

- 2

$- 2$ .

r = \frac{- 3 \pm \sqrt{9 + 8}}{2 \cdot 1}

$r = \frac{- 3 \pm \sqrt{9 + 8}}{2 \cdot 1}$

r = \frac{- 3 \pm \sqrt{9 + 8}}{2 \cdot 1}

$r = \frac{- 3 \pm \sqrt{9 + 8}}{2 \cdot 1}$

Step 3.1.3

Add

9

$9$ and

8

$8$ .

r = \frac{- 3 \pm \sqrt{17}}{2 \cdot 1}

$r = \frac{- 3 \pm \sqrt{17}}{2 \cdot 1}$

r = \frac{- 3 \pm \sqrt{17}}{2 \cdot 1}

$r = \frac{- 3 \pm \sqrt{17}}{2 \cdot 1}$

Step 3.2

Multiply

2

$2$ by

1

$1$ .

r = \frac{- 3 \pm \sqrt{17}}{2}

$r = \frac{- 3 \pm \sqrt{17}}{2}$

r = \frac{- 3 \pm \sqrt{17}}{2}

$r = \frac{- 3 \pm \sqrt{17}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

r = \frac{- 3 \pm \sqrt{17}}{2}

$r = \frac{- 3 \pm \sqrt{17}}{2}$

Decimal Form:

$r = 0.56155281 \dots, - 3.56155281 \dots$