Solve for q q^2+3q-28=0

$q^{2} + 3 q - 28 = 0$

Factor $q^{2} + 3 q - 28$ 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 $- 28$ and whose sum is $3$ .

$- 4, 7$

Write the factored form using these integers.

$(q - 4) (q + 7) = 0$

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

$q - 4 = 0$

$q + 7 = 0$

Set $q - 4$ equal to $0$ and solve for $q$ .

Set $q - 4$ equal to $0$ .

$q - 4 = 0$

Add $4$ to both sides of the equation.

$q = 4$

Set $q + 7$ equal to $0$ and solve for $q$ .

Set $q + 7$ equal to $0$ .

$q + 7 = 0$

Subtract $7$ from both sides of the equation.

$q = - 7$

The final solution is all the values that make $(q - 4) (q + 7) = 0$ true.

$q = 4, - 7$

Solve for q q^2+3q-28=0

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