What are the solutions to the equation (x+x+2)^2=x^2+(x+2)^2+336 ?

The solutions to the equation (x+x+2)^2=x^2+(x+2)^2+336 are x=12,x=-14

Find the zeros of (x+x+2)^2=x^2+(x+2)^2+336

The zeros of (x+x+2)^2=x^2+(x+2)^2+336 are x=12,x=-14

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + x + 2)^{2} = x^{2} + (x + 2)^{2} + 336

x = 12, x = - 14

Solution steps

(x + x + 2)^{2} = x^{2} + (x + 2)^{2} + 336

Expand

(x + x + 2)^{2} : 4 x^{2} + 8 x + 4

Expand

x^{2} + (x + 2)^{2} + 336 : 2 x^{2} + 4 x + 340

4 x^{2} + 8 x + 4 = 2 x^{2} + 4 x + 340

Move

340

to the left side

4 x^{2} + 8 x - 336 = 2 x^{2} + 4 x

Move

4 x

to the left side

4 x^{2} + 4 x - 336 = 2 x^{2}

Move

2 x^{2}

to the left side

2 x^{2} + 4 x - 336 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 336 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 336) = 52

x_{1, 2} = \frac{- 4 \pm 52}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 52}{2 \cdot 2}, x_{2} = \frac{- 4 - 52}{2 \cdot 2}

x = \frac{- 4 + 52}{2 \cdot 2} : 12

x = \frac{- 4 - 52}{2 \cdot 2} : - 14

The solutions to the quadratic equation are:

x = 12, x = - 14

(9 n + 1)^{2} = 9

4 x^{2} + 5 x - 26 x = \frac{13}{4}

- 15 - 7 x + 2 x^{2} = 0

2 x^{2} = - 8 x - 3

x^{2} + 2 x + 80 = 0

What are the solutions to the equation (x+x+2)^2=x^2+(x+2)^2+336 ?
The solutions to the equation (x+x+2)^2=x^2+(x+2)^2+336 are x=12,x=-14
Find the zeros of (x+x+2)^2=x^2+(x+2)^2+336
The zeros of (x+x+2)^2=x^2+(x+2)^2+336 are x=12,x=-14