What are the solutions to the equation x^2+(x-2)^2=10^2 ?

The solutions to the equation x^2+(x-2)^2=10^2 are x=8,x=-6

Find the zeros of x^2+(x-2)^2=10^2

The zeros of x^2+(x-2)^2=10^2 are x=8,x=-6

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x^{2} + (x - 2)^{2} = 1 0^{2}

x = 8, x = - 6

Solution steps

x^{2} + (x - 2)^{2} = 1 0^{2}

Expand

x^{2} + (x - 2)^{2} : 2 x^{2} - 4 x + 4

Expand

1 0^{2} : 100

2 x^{2} - 4 x + 4 = 100

Move

100

to the left side

2 x^{2} - 4 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 96 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 96) = 28

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 28}{2 \cdot 2} : 8

x = \frac{- ( - 4 ) - 28}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

x = 8, x = - 6

- 4.6 p - 6.3 p + 3.9 = - 9.18

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

2 x^{2} - 3 x + 13 = 0

b^{2} - 2 b + 1

4 x^{2} - 7 x + 10 = 0

What are the solutions to the equation x^2+(x-2)^2=10^2 ?
The solutions to the equation x^2+(x-2)^2=10^2 are x=8,x=-6
Find the zeros of x^2+(x-2)^2=10^2
The zeros of x^2+(x-2)^2=10^2 are x=8,x=-6