What are the solutions to the equation x^2+(x+1)^2+(x+2)^2+(x+3)^2=54 ?

The solutions to the equation x^2+(x+1)^2+(x+2)^2+(x+3)^2=54 are x=2,x=-5

Find the zeros of x^2+(x+1)^2+(x+2)^2+(x+3)^2=54

The zeros of x^2+(x+1)^2+(x+2)^2+(x+3)^2=54 are x=2,x=-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + (x + 1)^{2} + (x + 2)^{2} + (x + 3)^{2} = 54

x = 2, x = - 5

Solution steps

x^{2} + (x + 1)^{2} + (x + 2)^{2} + (x + 3)^{2} = 54

Expand

x^{2} + (x + 1)^{2} + (x + 2)^{2} + (x + 3)^{2} : 4 x^{2} + 12 x + 14

4 x^{2} + 12 x + 14 = 54

Move

54

to the left side

4 x^{2} + 12 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 40 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 40) = 28

x_{1, 2} = \frac{- 12 \pm 28}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 28}{2 \cdot 4}, x_{2} = \frac{- 12 - 28}{2 \cdot 4}

x = \frac{- 12 + 28}{2 \cdot 4} : 2

x = \frac{- 12 - 28}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

\frac{8 x ^{7} x}{12 x ^{5} y z ^{- 2}}

x^{2} - 5 x - 6

\frac{2}{( 3 ) ^{\frac{3}{2}}}

\frac{5 x - 8}{7} = - 5 x + 16

8 m^{6} - 2 m^{2}

What are the solutions to the equation x^2+(x+1)^2+(x+2)^2+(x+3)^2=54 ?
The solutions to the equation x^2+(x+1)^2+(x+2)^2+(x+3)^2=54 are x=2,x=-5
Find the zeros of x^2+(x+1)^2+(x+2)^2+(x+3)^2=54
The zeros of x^2+(x+1)^2+(x+2)^2+(x+3)^2=54 are x=2,x=-5