Solve for y 2|y+3|^2-9|y+3|-5=0

$2 {| y + 3 |}^{2} - 9 | y + 3 | - 5 = 0$

Factor the left side of the equation.

Let $u = | y + 3 |$ . Substitute $u$ for all occurrences of $| y + 3 |$ .

$2 u^{2} - 9 u - 5 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 u$ .

$2 u^{2} - 9 u - 5 = 0$

Rewrite $- 9$ as $1$ plus $- 10$

$2 u^{2} + (1 - 10) u - 5 = 0$

Apply the distributive property.

$2 u^{2} + 1 u - 10 u - 5 = 0$

Multiply $u$ by $1$ .

$2 u^{2} + u - 10 u - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} + u) - 10 u - 5 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (2 u + 1) - 5 (2 u + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

$(2 u + 1) (u - 5) = 0$

Replace all occurrences of $u$ with $| y + 3 |$ .

$(2 | y + 3 | + 1) (| y + 3 | - 5) = 0$

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

$2 | y + 3 | + 1 = 0$

$| y + 3 | - 5 = 0$

Set $2 | y + 3 | + 1$ equal to $0$ and solve for $y$ .

Set $2 | y + 3 | + 1$ equal to $0$ .

$2 | y + 3 | + 1 = 0$

Solve $2 | y + 3 | + 1 = 0$ for $y$ .

Subtract $1$ from both sides of the equation.

$2 | y + 3 | = - 1$

Divide each term in $2 | y + 3 | = - 1$ by $2$ and simplify.

Divide each term in $2 | y + 3 | = - 1$ by $2$ .

$\frac{2 | y + 3 |}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 | y + 3 |}{2} = \frac{- 1}{2}$

Divide $| y + 3 |$ by $1$ .

$| y + 3 | = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$| y + 3 | = - \frac{1}{2}$

There is no value of $y$ that makes the equation be true since an absolute value can never be negative.

No solution

Set $| y + 3 | - 5$ equal to $0$ and solve for $y$ .

Set $| y + 3 | - 5$ equal to $0$ .

$| y + 3 | - 5 = 0$

Solve $| y + 3 | - 5 = 0$ for $y$ .

Add $5$ to both sides of the equation.

$| y + 3 | = 5$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y + 3 = \pm 5$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y + 3 = 5$

Move all terms not containing $y$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$y = 5 - 3$

Subtract $3$ from $5$ .

$y = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$y + 3 = - 5$

Move all terms not containing $y$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$y = - 5 - 3$

Subtract $3$ from $- 5$ .

$y = - 8$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 2, - 8$

The final solution is all the values that make $(2 | y + 3 | + 1) (| y + 3 | - 5) = 0$ true.

$y = 2, - 8$

Solve for y 2|y+3|^2-9|y+3|-5=0

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