Pre-Algebra Examples

$5 r^{2} = - 2 r + 20$

Step 1

Add $2 r$ to both sides of the equation.

$5 r^{2} + 2 r = 20$

Step 2

Subtract $20$ from both sides of the equation.

$5 r^{2} + 2 r - 20 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 5$ , $b = 2$ , and $c = - 20$ into the quadratic formula and solve for $r$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (5 \cdot - 20)}}{2 \cdot 5}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

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

$r = \frac{- 2 \pm \sqrt{4 - 4 \cdot 5 \cdot - 20}}{2 \cdot 5}$

Step 5.1.2

Multiply $- 4 \cdot 5 \cdot - 20$ .

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Step 5.1.2.1

Multiply $- 4$ by $5$ .

$r = \frac{- 2 \pm \sqrt{4 - 20 \cdot - 20}}{2 \cdot 5}$

Step 5.1.2.2

Multiply $- 20$ by $- 20$ .

$r = \frac{- 2 \pm \sqrt{4 + 400}}{2 \cdot 5}$

Step 5.1.3

Add $4$ and $400$ .

$r = \frac{- 2 \pm \sqrt{404}}{2 \cdot 5}$

Step 5.1.4

Rewrite $404$ as $2^{2} \cdot 101$ .

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Step 5.1.4.1

Factor $4$ out of $404$ .

$r = \frac{- 2 \pm \sqrt{4 (101)}}{2 \cdot 5}$

Step 5.1.4.2

Rewrite $4$ as $2^{2}$ .

$r = \frac{- 2 \pm \sqrt{2^{2} \cdot 101}}{2 \cdot 5}$

Step 5.1.5

Pull terms out from under the radical.

$r = \frac{- 2 \pm 2 \sqrt{101}}{2 \cdot 5}$

Step 5.2

Multiply $2$ by $5$ .

$r = \frac{- 2 \pm 2 \sqrt{101}}{10}$

Step 5.3

Simplify $\frac{- 2 \pm 2 \sqrt{101}}{10}$ .

$r = \frac{- 1 \pm \sqrt{101}}{5}$

Step 6

The final answer is the combination of both solutions.

$r = - \frac{1 - \sqrt{101}}{5}, - \frac{1 + \sqrt{101}}{5}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{1 - \sqrt{101}}{5}, - \frac{1 + \sqrt{101}}{5}$

Decimal Form:

$r = 1.80997512 \dots, - 2.20997512 \dots$