Solve for q q^2+7q-18=0

$q^{2} + 7 q - 18 = 0$

Step 1

Factor $q^{2} + 7 q - 18$ 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 $- 18$ and whose sum is $7$ .

$- 2, 9$

Write the factored form using these integers.

$(q - 2) (q + 9) = 0$

Step 2

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

$q - 2 = 0$

$q + 9 = 0$

Step 3

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

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

$q - 2 = 0$

Add $2$ to both sides of the equation.

$q = 2$

Step 4

Set $q + 9$ equal to $0$ and solve for $q$ .

Set $q + 9$ equal to $0$ .

$q + 9 = 0$

Subtract $9$ from both sides of the equation.

$q = - 9$

Step 5

The final solution is all the values that make $(q - 2) (q + 9) = 0$ true.

$q = 2, - 9$

Solve for q q^2+7q-18=0

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