What are the solutions to the equation x^2+(x+1)^2=80 ?

The solutions to the equation x^2+(x+1)^2=80 are x=(-1+sqrt(159))/2 ,x=-(1+sqrt(159))/2

Find the zeros of x^2+(x+1)^2=80

The zeros of x^2+(x+1)^2=80 are x=(-1+sqrt(159))/2 ,x=-(1+sqrt(159))/2

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x^{2} + (x + 1)^{2} = 80

x = \frac{- 1 + 159}{2}, x = - \frac{1 + 159}{2}

Decimal Notation

x = 5.80476 \dots, x = - 6.80476 \dots

Solution steps

x^{2} + (x + 1)^{2} = 80

Expand

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

2 x^{2} + 2 x + 1 = 80

Move

80

to the left side

2 x^{2} + 2 x - 79 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 79 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 79) = 2159

x_{1, 2} = \frac{- 2 \pm 2 159}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 159}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 159}{2 \cdot 2}

x = \frac{- 2 + 2 159}{2 \cdot 2} : \frac{- 1 + 159}{2}

x = \frac{- 2 - 2 159}{2 \cdot 2} : - \frac{1 + 159}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 159}{2}, x = - \frac{1 + 159}{2}

x + 9 \leq 2

c^{2} + 5 c = 14

a^{4} - 5 a^{2} - 36

\frac{2}{5} x = - \frac{1}{10}

What are the solutions to the equation x^2+(x+1)^2=80 ?
The solutions to the equation x^2+(x+1)^2=80 are x=(-1+sqrt(159))/2 ,x=-(1+sqrt(159))/2
Find the zeros of x^2+(x+1)^2=80
The zeros of x^2+(x+1)^2=80 are x=(-1+sqrt(159))/2 ,x=-(1+sqrt(159))/2