What are the solutions to the equation x^2+(x+1)^2+(x+2)^2=1454 ?

The solutions to the equation x^2+(x+1)^2+(x+2)^2=1454 are x=21,x=-23

Find the zeros of x^2+(x+1)^2+(x+2)^2=1454

The zeros of x^2+(x+1)^2+(x+2)^2=1454 are x=21,x=-23

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x^{2} + (x + 1)^{2} + (x + 2)^{2} = 1454

x = 21, x = - 23

Solution steps

x^{2} + (x + 1)^{2} + (x + 2)^{2} = 1454

Expand

x^{2} + (x + 1)^{2} + (x + 2)^{2} : 3 x^{2} + 6 x + 5

3 x^{2} + 6 x + 5 = 1454

Move

1454

to the left side

3 x^{2} + 6 x - 1449 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 1449 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 1449) = 132

x_{1, 2} = \frac{- 6 \pm 132}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 132}{2 \cdot 3}, x_{2} = \frac{- 6 - 132}{2 \cdot 3}

x = \frac{- 6 + 132}{2 \cdot 3} : 21

x = \frac{- 6 - 132}{2 \cdot 3} : - 23

The solutions to the quadratic equation are:

x = 21, x = - 23

3 x^{2} - 37 = 0

3 x^{2} + 3 x + 4 = 0

3 x^{2} + 3 x + 5 = 0

3 x^{2} + 10 x - 2 = 0

5 x^{2} - 25 x - 180 = 0

What are the solutions to the equation x^2+(x+1)^2+(x+2)^2=1454 ?
The solutions to the equation x^2+(x+1)^2+(x+2)^2=1454 are x=21,x=-23
Find the zeros of x^2+(x+1)^2+(x+2)^2=1454
The zeros of x^2+(x+1)^2+(x+2)^2=1454 are x=21,x=-23