What are the solutions to the equation (x+x+1)/2 =(x+1)^2-300 ?

The solutions to the equation (x+x+1)/2 =(x+1)^2-300 are x=(-1+sqrt(1199))/2 ,x=-(1+sqrt(1199))/2

Find the zeros of (x+x+1)/2 =(x+1)^2-300

The zeros of (x+x+1)/2 =(x+1)^2-300 are x=(-1+sqrt(1199))/2 ,x=-(1+sqrt(1199))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{x + x + 1}{2} = (x + 1)^{2} - 300

x = \frac{- 1 + 1199}{2}, x = - \frac{1 + 1199}{2}

Decimal Notation

x = 16.81328 \dots, x = - 17.81328 \dots

Solution steps

\frac{x + x + 1}{2} = (x + 1)^{2} - 300

Multiply both sides by

2

2 x + 1 = 2 (x + 1)^{2} - 600

Expand

2 (x + 1)^{2} - 600 : 2 x^{2} + 4 x - 598

2 x + 1 = 2 x^{2} + 4 x - 598

Switch sides

2 x^{2} + 4 x - 598 = 2 x + 1

Move

1

to the left side

2 x^{2} + 4 x - 599 = 2 x

Move

2 x

to the left side

2 x^{2} + 2 x - 599 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 599 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 599) = 21199

x_{1, 2} = \frac{- 2 \pm 2 1199}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 1199}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 1199}{2 \cdot 2}

x = \frac{- 2 + 2 1199}{2 \cdot 2} : \frac{- 1 + 1199}{2}

x = \frac{- 2 - 2 1199}{2 \cdot 2} : - \frac{1 + 1199}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 1199}{2}, x = - \frac{1 + 1199}{2}

3 x^{2} = 12 x - 1

2 x^{2} + 3 x - 1 = 0

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

x, x^{2} + y^{2} + z^{2} + x + 8 y + m = 0

- y^{2} + 10 y = 24

What are the solutions to the equation (x+x+1)/2 =(x+1)^2-300 ?
The solutions to the equation (x+x+1)/2 =(x+1)^2-300 are x=(-1+sqrt(1199))/2 ,x=-(1+sqrt(1199))/2
Find the zeros of (x+x+1)/2 =(x+1)^2-300
The zeros of (x+x+1)/2 =(x+1)^2-300 are x=(-1+sqrt(1199))/2 ,x=-(1+sqrt(1199))/2