What are the solutions to the equation x^2+(2-x)^2=4 ?

The solutions to the equation x^2+(2-x)^2=4 are x=2,x=0

Find the zeros of x^2+(2-x)^2=4

The zeros of x^2+(2-x)^2=4 are x=2,x=0

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

x = 2, x = 0

Solution steps

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

Expand

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

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

Move

4

to the left side

2 x^{2} - 4 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 \cdot 0 = 4

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

Separate the solutions

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

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

x = \frac{- ( - 4 ) - 4}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

y, y^{2} = e^{2 t} + k

0.4 x^{2} + 0.242 x - 0.042 = 0

x + 2 (x - 3)^{2} + 2 = 3 (x + 2)^{2}

0 = - x^{2} + 6 x - 15

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

What are the solutions to the equation x^2+(2-x)^2=4 ?
The solutions to the equation x^2+(2-x)^2=4 are x=2,x=0
Find the zeros of x^2+(2-x)^2=4
The zeros of x^2+(2-x)^2=4 are x=2,x=0