Basic Math Examples

$r (r + 5) (2) = 90$

Step 1

Divide each term in $r (r + 5) \cdot 2 = 90$ by $2$ and simplify.

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Step 1.1

Divide each term in $r (r + 5) \cdot 2 = 90$ by $2$ .

$\frac{r (r + 5) \cdot 2}{2} = \frac{90}{2}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $2$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{r (r + 5) \cdot 2}{2} = \frac{90}{2}$

Step 1.2.1.2

Divide $r (r + 5)$ by $1$ .

$r (r + 5) = \frac{90}{2}$

Step 1.2.2

Apply the distributive property.

$r \cdot r + r \cdot 5 = \frac{90}{2}$

Step 1.2.3

Simplify the expression.

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Step 1.2.3.1

Multiply $r$ by $r$ .

$r^{2} + r \cdot 5 = \frac{90}{2}$

Step 1.2.3.2

Move $5$ to the left of $r$ .

$r^{2} + 5 r = \frac{90}{2}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $90$ by $2$ .

$r^{2} + 5 r = 45$

Step 2

Subtract $45$ from both sides of the equation.

$r^{2} + 5 r - 45 = 0$

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 $a = 1$ , $b = 5$ , and $c = - 45$ into the quadratic formula and solve for $r$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 45)}}{2 \cdot 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $5$ to the power of $2$ .

$r = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 45}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot - 45$ .

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$r = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 45}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 45$ .

$r = \frac{- 5 \pm \sqrt{25 + 180}}{2 \cdot 1}$

Step 5.1.3

Add $25$ and $180$ .

$r = \frac{- 5 \pm \sqrt{205}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$r = \frac{- 5 \pm \sqrt{205}}{2}$

Step 6

The final answer is the combination of both solutions.

$r = - \frac{5 - \sqrt{205}}{2}, - \frac{5 + \sqrt{205}}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{5 - \sqrt{205}}{2}, - \frac{5 + \sqrt{205}}{2}$

Decimal Form:

$r = 4.65891053 \dots, - 9.65891053 \dots$