What are the solutions to the equation 2x-(x+1)^2=3x^2-6 ?

The solutions to the equation 2x-(x+1)^2=3x^2-6 are x=(sqrt(5))/2 ,x=-(sqrt(5))/2

Find the zeros of 2x-(x+1)^2=3x^2-6

The zeros of 2x-(x+1)^2=3x^2-6 are x=(sqrt(5))/2 ,x=-(sqrt(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x - (x + 1)^{2} = 3 x^{2} - 6

x = \frac{5}{2}, x = - \frac{5}{2}

Decimal Notation

x = 1.11803 \dots, x = - 1.11803 \dots

Solution steps

2 x - (x + 1)^{2} = 3 x^{2} - 6

Expand

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

- x^{2} - 1 = 3 x^{2} - 6

Move

1

to the right side

- x^{2} = 3 x^{2} - 5

Move

3 x^{2}

to the left side

- 4 x^{2} = - 5

Divide both sides by

- 4

x^{2} = \frac{5}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{5}{4}, x = - \frac{5}{4}

\frac{5}{4} = \frac{5}{2}

- \frac{5}{4} = - \frac{5}{2}

x = \frac{5}{2}, x = - \frac{5}{2}

2 q^{2} - 13 q + 20 = 0

3 a^{2} - 3 a - 6

5 x^{2} + 3 x = 0

\frac{1}{0.1 i + 1}

2 (x + 6) > 2 x + 4

What are the solutions to the equation 2x-(x+1)^2=3x^2-6 ?
The solutions to the equation 2x-(x+1)^2=3x^2-6 are x=(sqrt(5))/2 ,x=-(sqrt(5))/2
Find the zeros of 2x-(x+1)^2=3x^2-6
The zeros of 2x-(x+1)^2=3x^2-6 are x=(sqrt(5))/2 ,x=-(sqrt(5))/2