Solve for q 3q^2+q-14=0

$3 q^{2} + q - 14 = 0$

Step 1

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 $a \cdot c = 3 \cdot - 14 = - 42$ and whose sum is $b = 1$ .

Multiply by $1$ .

$3 q^{2} + 1 q - 14 = 0$

Rewrite $1$ as $- 6$ plus $7$

$3 q^{2} + (- 6 + 7) q - 14 = 0$

Apply the distributive property.

$3 q^{2} - 6 q + 7 q - 14 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 q^{2} - 6 q) + 7 q - 14 = 0$

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

$3 q (q - 2) + 7 (q - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $q - 2$ .

$(q - 2) (3 q + 7) = 0$

Step 2

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

$q - 2 = 0$

$3 q + 7 = 0$

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 $3 q + 7$ equal to $0$ and solve for $q$ .

Set $3 q + 7$ equal to $0$ .

$3 q + 7 = 0$

Solve $3 q + 7 = 0$ for $q$ .

Subtract $7$ from both sides of the equation.

$3 q = - 7$

Divide each term in $3 q = - 7$ by $3$ and simplify.

Divide each term in $3 q = - 7$ by $3$ .

$\frac{3 q}{3} = \frac{- 7}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 q}{3} = \frac{- 7}{3}$

Divide $q$ by $1$ .

$q = \frac{- 7}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$q = - \frac{7}{3}$

Step 5

The final solution is all the values that make $(q - 2) (3 q + 7) = 0$ true.

$q = 2, - \frac{7}{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$q = 2, - \frac{7}{3}$

Decimal Form:

$q = 2, - 2.\overline{3}$

Mixed Number Form:

$q = 2, - 2 \frac{1}{3}$

Solve for q 3q^2+q-14=0

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