What is the answer to w(4w+6)+2w=2(2w^2+7w-3) ?

The answer to w(4w+6)+2w=2(2w^2+7w-3) is w=1

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w(4w+6)+2w=2(2w^2+7w-3)

w (4 w + 6) + 2 w = 2 (2 w^{2} + 7 w - 3)

w = 1

Solution steps

w (4 w + 6) + 2 w = 2 (2 w^{2} + 7 w - 3)

Expand

w (4 w + 6) : 4 w^{2} + 6 w

4 w^{2} + 6 w + 2 w = 2 (2 w^{2} + 7 w - 3)

Add similar elements:

6 w + 2 w = 8 w

4 w^{2} + 8 w = 2 (2 w^{2} + 7 w - 3)

Expand

2 (2 w^{2} + 7 w - 3) : 4 w^{2} + 14 w - 6

4 w^{2} + 8 w = 4 w^{2} + 14 w - 6

Subtract

4 w^{2} + 14 w

from both sides

4 w^{2} + 8 w - (4 w^{2} + 14 w) = 4 w^{2} + 14 w - 6 - (4 w^{2} + 14 w)

Simplify

- 6 w = - 6

Divide both sides by

- 6

w = 1

\frac{3}{25} \cdot \frac{2}{5}

3 x^{2} + 45 = 0

(x + 2)^{2}

6 p - 2 - 3 p = 16

- 3 (x + 5) > 39

What is the answer to w(4w+6)+2w=2(2w^2+7w-3) ?
The answer to w(4w+6)+2w=2(2w^2+7w-3) is w=1
Solve for w: w(4w+6)+2w=2(2w^2+7w-3)
The solution is w=1