Solve for w 3w^2-4w=A

$3 w^{2} - 4 w = A$

Step 1

Subtract $A$ from both sides of the equation.

$3 w^{2} - 4 w - A = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

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

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

Step 4

Simplify.

Simplify the numerator.

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 3 \cdot - 1 A$ .

Factor $- 4$ out of ${(- 4)}^{2}$ .

$w = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 3 \cdot (- 1 A)}}{2 \cdot 3}$

Factor $- 4$ out of $- 4 \cdot 3 \cdot - 1 A$ .

$w = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (3 \cdot (- 1 A))}}{2 \cdot 3}$

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (3 \cdot - 1 A)$ .

$w = \frac{4 \pm \sqrt{- 4 (- 4 + 3 \cdot (- 1 A))}}{2 \cdot 3}$

Multiply $3$ by $- 1$ .

$w = \frac{4 \pm \sqrt{- 4 (- 4 - 3 A)}}{2 \cdot 3}$

Rewrite $- 4 (- 4 - 3 A)$ as $2^{2} \cdot (- 1 (- 4 - 3 A))$ .

Factor $4$ out of $- 4$ .

$w = \frac{4 \pm \sqrt{4 (- 1) (- 4 - 3 A)}}{2 \cdot 3}$

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

$w = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 4 - 3 A))}}{2 \cdot 3}$

Add parentheses.

$w = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 4 - 3 A))}}{2 \cdot 3}$

Pull terms out from under the radical.

$w = \frac{4 \pm 2 \sqrt{- 1 (- 4 - 3 A)}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$w = \frac{4 \pm 2 \sqrt{- 1 (- 4 - 3 A)}}{6}$

Simplify $\frac{4 \pm 2 \sqrt{- 1 (- 4 - 3 A)}}{6}$ .

$w = \frac{2 \pm \sqrt{- 1 (- 4 - 3 A)}}{3}$

Step 5

The final answer is the combination of both solutions.

$w = \frac{2 + \sqrt{- 1 (- 4 - 3 A)}}{3}$

$w = \frac{2 - \sqrt{- 1 (- 4 - 3 A)}}{3}$