What are the solutions to the equation x+3=(2-x)^2 ?

The solutions to the equation x+3=(2-x)^2 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2

Find the zeros of x+3=(2-x)^2

The zeros of x+3=(2-x)^2 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2

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

x = \frac{5 + 21}{2}, x = \frac{5 - 21}{2}

Solution steps

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

Expand

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

x + 3 = 4 - 4 x + x^{2}

Switch sides

4 - 4 x + x^{2} = x + 3

Move

3

to the left side

x^{2} - 4 x + 1 = x

Move

x

to the left side

x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 1 = 21

x_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 1}

x = \frac{- ( - 5 ) + 21}{2 \cdot 1} : \frac{5 + 21}{2}

x = \frac{- ( - 5 ) - 21}{2 \cdot 1} : \frac{5 - 21}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 21}{2}, x = \frac{5 - 21}{2}

x^{2} - 4 x + 3 x - 2 = 28

6 a^{2} + 7 a - 3 = 0

12 x^{2} + 40 x - 32 = 0

2 x^{2} = x + 4

3 x + x^{2} - 10 = 5

What are the solutions to the equation x+3=(2-x)^2 ?
The solutions to the equation x+3=(2-x)^2 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2
Find the zeros of x+3=(2-x)^2
The zeros of x+3=(2-x)^2 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2