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

Предыдущий
вопрос Следующий
вопрос

To solve this equation, first distribute the 2 on the left side:

2x - 6 - 2x^2 = x - (2x+1)^2

Next, square the binomial on the right side:

2x - 6 - 2x^2 = x - (4x^2 + 4x + 1)

Next, distribute the negative sign inside the parentheses:

2x - 6 - 2x^2 = x - 4x^2 - 4x - 1

Combine like terms on both sides:

-2x^2 + 2x - 6 = -3x^2 - 3x - 1

To continue solving for x, we can move all terms to one side of the equation and set it equal to zero:

0 = -x^2 - 5x + 5

Next, we can factor or use the quadratic formula to solve for x. Let's use the quadratic formula:

x = (-(-5) ± sqrt((-5)^2 - 4(1)(5))) / 2(1)
x = (5 ± sqrt(25 - 20)) / 2
x = (5 ± sqrt(5)) / 2

Therefore, the solutions for x are:

x = (5 + sqrt(5)) / 2
x = (5 - sqrt(5)) / 2

These are the solutions to the equation 2(x-3)-2x^2=x-(2x+1)^2.