Solve for x (x+3)^2+(x+3)-2=0

${(x + 3)}^{2} + (x + 3) - 2 = 0$

Step 1

Subtract $2$ from $3$ .

${(x + 3)}^{2} + x + 1 = 0$

Step 2

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u^{2} + 7 u + 10 = 0$

Factor $u^{2} + 7 u + 10$ using the AC method.

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 $7$ .

$2, 5$

Write the factored form using these integers.

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

Replace all occurrences of $u$ with $x$ .

$(x + 2) (x + 5) = 0$

Step 3

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

$x + 2 = 0$

$x + 5 = 0$

Step 4

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

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

$x = - 2, - 5$

Solve for x (x+3)^2+(x+3)-2=0

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