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

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

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

The zeros of ((x-2)^2)/2+2(2-x)=4 are x=2(2+sqrt(3)),x=2(2-sqrt(3))

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

x = 2 (2 + 3), x = 2 (2 - 3)

Decimal Notation

x = 7.46410 \dots, x = 0.53589 \dots

Solution steps

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

Multiply both sides by

2

(x - 2)^{2} + 4 (2 - x) = 8

Expand

(x - 2)^{2} + 4 (2 - x) : x^{2} - 8 x + 12

x^{2} - 8 x + 12 = 8

Move

8

to the left side

x^{2} - 8 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 4 = 43

x_{1, 2} = \frac{- ( - 8 ) \pm 4 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 3}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 4 3}{2 \cdot 1}

x = \frac{- ( - 8 ) + 4 3}{2 \cdot 1} : 2 (2 + 3)

x = \frac{- ( - 8 ) - 4 3}{2 \cdot 1} : 2 (2 - 3)

The solutions to the quadratic equation are:

x = 2 (2 + 3), x = 2 (2 - 3)

3 x^{2} - 12 x = 6

1 x^{2} + 2 x - 8 = 0

y, (4 - x) y^{2} = x^{3}

3 x^{2} + 18 x + 20

x^{2} + 10 + 9 = 0

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