Basic Math Examples

$3 r (r + 3) = 30$

Step 1

Divide each term in $3 r (r + 3) = 30$ by $3$ and simplify.

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Step 1.1

Divide each term in $3 r (r + 3) = 30$ by $3$ .

$\frac{3 r (r + 3)}{3} = \frac{30}{3}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $3$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{3 r (r + 3)}{3} = \frac{30}{3}$

Step 1.2.1.2

Divide $r (r + 3)$ by $1$ .

$r (r + 3) = \frac{30}{3}$

Step 1.2.2

Apply the distributive property.

$r \cdot r + r \cdot 3 = \frac{30}{3}$

Step 1.2.3

Simplify the expression.

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Step 1.2.3.1

Multiply $r$ by $r$ .

$r^{2} + r \cdot 3 = \frac{30}{3}$

Step 1.2.3.2

Move $3$ to the left of $r$ .

$r^{2} + 3 r = \frac{30}{3}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $30$ by $3$ .

$r^{2} + 3 r = 10$

Step 2

Subtract $10$ from both sides of the equation.

$r^{2} + 3 r - 10 = 0$

Step 3

Factor $r^{2} + 3 r - 10$ using the AC method.

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Step 3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $3$ .

$- 2, 5$

Step 3.2

Write the factored form using these integers.

$(r - 2) (r + 5) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r - 2 = 0$

$r + 5 = 0$

Step 5

Set $r - 2$ equal to $0$ and solve for $r$ .

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Step 5.1

Set $r - 2$ equal to $0$ .

$r - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$r = 2$

Step 6

Set $r + 5$ equal to $0$ and solve for $r$ .

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Step 6.1

Set $r + 5$ equal to $0$ .

$r + 5 = 0$

Step 6.2

Subtract $5$ from both sides of the equation.

$r = - 5$

Step 7

The final solution is all the values that make $(r - 2) (r + 5) = 0$ true.

$r = 2, - 5$