What are the solutions to the equation x^2+(x+1)^2+(x+2)^2=110 ?

The solutions to the equation x^2+(x+1)^2+(x+2)^2=110 are x=5,x=-7

Find the zeros of x^2+(x+1)^2+(x+2)^2=110

The zeros of x^2+(x+1)^2+(x+2)^2=110 are x=5,x=-7

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x^{2} + (x + 1)^{2} + (x + 2)^{2} = 110

x = 5, x = - 7

Solution steps

x^{2} + (x + 1)^{2} + (x + 2)^{2} = 110

Expand

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

3 x^{2} + 6 x + 5 = 110

Move

110

to the left side

3 x^{2} + 6 x - 105 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 105 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 105) = 36

x_{1, 2} = \frac{- 6 \pm 36}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 36}{2 \cdot 3}, x_{2} = \frac{- 6 - 36}{2 \cdot 3}

x = \frac{- 6 + 36}{2 \cdot 3} : 5

x = \frac{- 6 - 36}{2 \cdot 3} : - 7

The solutions to the quadratic equation are:

x = 5, x = - 7

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

70840 - 188 d - 2.35 d^{2} = 1228.8 d + 30.72 d^{2}

u^{2} - 16 u

n (n + 1) = 0.3 n (n) + 300

12 x^{2} + 29 x + 14 = 0

What are the solutions to the equation x^2+(x+1)^2+(x+2)^2=110 ?
The solutions to the equation x^2+(x+1)^2+(x+2)^2=110 are x=5,x=-7
Find the zeros of x^2+(x+1)^2+(x+2)^2=110
The zeros of x^2+(x+1)^2+(x+2)^2=110 are x=5,x=-7