Solve for q 3q^2=4-6q

$3 q^{2} = 4 - 6 q$

Step 1

Add $6 q$ to both sides of the equation.

$3 q^{2} + 6 q = 4$

Step 2

Subtract $4$ from both sides of the equation.

$3 q^{2} + 6 q - 4 = 0$

Step 3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4

Substitute the values $a = 3$ , $b = 6$ , and $c = - 4$ into the quadratic formula and solve for $q$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (3 \cdot - 4)}}{2 \cdot 3}$

Step 5

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$q = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot - 4}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 4$ .

Multiply $- 4$ by $3$ .

$q = \frac{- 6 \pm \sqrt{36 - 12 \cdot - 4}}{2 \cdot 3}$

Multiply $- 12$ by $- 4$ .

$q = \frac{- 6 \pm \sqrt{36 + 48}}{2 \cdot 3}$

Add $36$ and $48$ .

$q = \frac{- 6 \pm \sqrt{84}}{2 \cdot 3}$

Rewrite $84$ as $2^{2} \cdot 21$ .

Factor $4$ out of $84$ .

$q = \frac{- 6 \pm \sqrt{4 (21)}}{2 \cdot 3}$

Rewrite $4$ as $2^{2}$ .

$q = \frac{- 6 \pm \sqrt{2^{2} \cdot 21}}{2 \cdot 3}$

Pull terms out from under the radical.

$q = \frac{- 6 \pm 2 \sqrt{21}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$q = \frac{- 6 \pm 2 \sqrt{21}}{6}$

Simplify $\frac{- 6 \pm 2 \sqrt{21}}{6}$ .

$q = \frac{- 3 \pm \sqrt{21}}{3}$

Step 6

The final answer is the combination of both solutions.

$q = - \frac{3 - \sqrt{21}}{3}, - \frac{3 + \sqrt{21}}{3}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$q = - \frac{3 - \sqrt{21}}{3}, - \frac{3 + \sqrt{21}}{3}$

Decimal Form:

$q = 0.52752523 \dots, - 2.52752523 \dots$