Solve Using the Quadratic Formula 1/12y^2=-y-3

$\frac{1}{12} y^{2} = - y - 3$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Combine $\frac{1}{12}$ and $y^{2}$ .

$\frac{y^{2}}{12} = - y - 3$

Move all the expressions to the left side of the equation.

Add $y$ to both sides of the equation.

$\frac{y^{2}}{12} + y = - 3$

Add $3$ to both sides of the equation.

$\frac{y^{2}}{12} + y + 3 = 0$

Step 2

Multiply through by the least common denominator $12$ , then simplify.

Apply the distributive property.

$12 (\frac{y^{2}}{12}) + 12 y + 12 \cdot 3 = 0$

Simplify.

Cancel the common factor of $12$ .

Cancel the common factor.

$12 (\frac{y^{2}}{12}) + 12 y + 12 \cdot 3 = 0$

Rewrite the expression.

$y^{2} + 12 y + 12 \cdot 3 = 0$

Multiply $12$ by $3$ .

$y^{2} + 12 y + 36 = 0$

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 $a = 1$ , $b = 12$ , and $c = 36$ into the quadratic formula and solve for $y$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot 36)}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$y = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 36$ .

Multiply $- 4$ by $1$ .

$y = \frac{- 12 \pm \sqrt{144 - 4 \cdot 36}}{2 \cdot 1}$

Multiply $- 4$ by $36$ .

$y = \frac{- 12 \pm \sqrt{144 - 144}}{2 \cdot 1}$

Subtract $144$ from $144$ .

$y = \frac{- 12 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$y = \frac{- 12 \pm \sqrt{0^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \frac{- 12 \pm 0}{2 \cdot 1}$

$- 12$ plus or minus $0$ is $- 12$ .

$y = \frac{- 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 12}{2}$

Divide $- 12$ by $2$ .

$y = - 6$

Step 6

The final answer is the combination of both solutions.

$y = - 6$ Double roots

Solve Using the Quadratic Formula 1/12y^2=-y-3

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