What are the solutions to the equation 25=x^2+(x+1)^2 ?

The solutions to the equation 25=x^2+(x+1)^2 are x=3,x=-4

Find the zeros of 25=x^2+(x+1)^2

The zeros of 25=x^2+(x+1)^2 are x=3,x=-4

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

x = 3, x = - 4

Solution steps

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

Expand

x^{2} + (x + 1)^{2} : 2 x^{2} + 2 x + 1

25 = 2 x^{2} + 2 x + 1

Switch sides

2 x^{2} + 2 x + 1 = 25

Move

25

to the left side

2 x^{2} + 2 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 24 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 24) = 14

x_{1, 2} = \frac{- 2 \pm 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 14}{2 \cdot 2}, x_{2} = \frac{- 2 - 14}{2 \cdot 2}

x = \frac{- 2 + 14}{2 \cdot 2} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 4

9 x^{2} + 5 = 6

- 2 x^{2} - 7 x + 13 = 0

x^{2} = 11664

12 x^{2} - 92 x + 130 = 0

6 p^{2} + 7 p - 20 = 0

What are the solutions to the equation 25=x^2+(x+1)^2 ?
The solutions to the equation 25=x^2+(x+1)^2 are x=3,x=-4
Find the zeros of 25=x^2+(x+1)^2
The zeros of 25=x^2+(x+1)^2 are x=3,x=-4