Basic Math Examples

${(y + 10)}^{2} = 2 y + 19$

Step 1

Move all terms containing $y$ to the left side of the equation.

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Step 1.1

Subtract $2 y$ from both sides of the equation.

${(y + 10)}^{2} - 2 y = 19$

Step 1.2

Simplify each term.

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Step 1.2.1

Rewrite ${(y + 10)}^{2}$ as $(y + 10) (y + 10)$ .

$(y + 10) (y + 10) - 2 y = 19$

Step 1.2.2

Expand $(y + 10) (y + 10)$ using the FOIL Method.

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Step 1.2.2.1

Apply the distributive property.

$y (y + 10) + 10 (y + 10) - 2 y = 19$

Step 1.2.2.2

Apply the distributive property.

$y \cdot y + y \cdot 10 + 10 (y + 10) - 2 y = 19$

Step 1.2.2.3

Apply the distributive property.

$y \cdot y + y \cdot 10 + 10 y + 10 \cdot 10 - 2 y = 19$

Step 1.2.3

Simplify and combine like terms.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Multiply $y$ by $y$ .

$y^{2} + y \cdot 10 + 10 y + 10 \cdot 10 - 2 y = 19$

Step 1.2.3.1.2

Move $10$ to the left of $y$ .

$y^{2} + 10 \cdot y + 10 y + 10 \cdot 10 - 2 y = 19$

Step 1.2.3.1.3

Multiply $10$ by $10$ .

$y^{2} + 10 y + 10 y + 100 - 2 y = 19$

Step 1.2.3.2

Add $10 y$ and $10 y$ .

$y^{2} + 20 y + 100 - 2 y = 19$

Step 1.3

Subtract $2 y$ from $20 y$ .

$y^{2} + 18 y + 100 = 19$

Step 2

Subtract $19$ from both sides of the equation.

$y^{2} + 18 y + 100 - 19 = 0$

Step 3

Subtract $19$ from $100$ .

$y^{2} + 18 y + 81 = 0$

Step 4

Factor using the perfect square rule.

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Step 4.1

Rewrite $81$ as $9^{2}$ .

$y^{2} + 18 y + 9^{2} = 0$

Step 4.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 y = 2 \cdot y \cdot 9$

Step 4.3

Rewrite the polynomial.

$y^{2} + 2 \cdot y \cdot 9 + 9^{2} = 0$

Step 4.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 9$ .

${(y + 9)}^{2} = 0$

Step 5

Set the $y + 9$ equal to $0$ .

$y + 9 = 0$

Step 6

Subtract $9$ from both sides of the equation.

$y = - 9$