What is the answer to (2x-1)(2x+3)=2(2x^2-x+1) ?

The answer to (2x-1)(2x+3)=2(2x^2-x+1) is x= 5/6

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(2x-1)(2x+3)=2(2x^2-x+1)

(2 x - 1) (2 x + 3) = 2 (2 x^{2} - x + 1)

x = \frac{5}{6}

Decimal Notation

x = 0.83333 \dots

Solution steps

(2 x - 1) (2 x + 3) = 2 (2 x^{2} - x + 1)

Expand

(2 x - 1) (2 x + 3) : 4 x^{2} + 4 x - 3

Expand

2 (2 x^{2} - x + 1) : 4 x^{2} - 2 x + 2

4 x^{2} + 4 x - 3 = 4 x^{2} - 2 x + 2

Move

3

to the right side

4 x^{2} + 4 x = 4 x^{2} - 2 x + 5

Subtract

4 x^{2} - 2 x

from both sides

4 x^{2} + 4 x - (4 x^{2} - 2 x) = 4 x^{2} - 2 x + 5 - (4 x^{2} - 2 x)

Simplify

6 x = 5

Divide both sides by

6

x = \frac{5}{6}

p^{2} + p - 6 = 0

3 m + 58 = m^{2} + 12 m + 36

- (2 - 3 x) < 5 (4 - x)

- 3 (3 c + 5) = 2 (10 - c)

0.6 x - 0.3 = 0.1

What is the answer to (2x-1)(2x+3)=2(2x^2-x+1) ?
The answer to (2x-1)(2x+3)=2(2x^2-x+1) is x= 5/6
Solve for x: (2x-1)(2x+3)=2(2x^2-x+1)
The solution is x= 5/6