What are the solutions to the equation [-x-3]^2=5-2[-x] ?

The solutions to the equation [-x-3]^2=5-2[-x] are x=-2

Find the zeros of [-x-3]^2=5-2[-x]

The zeros of [-x-3]^2=5-2[-x] are x=-2

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[- x - 3]^{2} = 5 - 2 [- x]

x = - 2

Solution steps

[- x - 3]^{2} = 5 - 2 [- x]

Expand

(- x - 3)^{2} : x^{2} + 6 x + 9

Expand

5 - 2 (- x) : 5 + 2 x

x^{2} + 6 x + 9 = 5 + 2 x

Move

2 x

to the left side

x^{2} + 4 x + 9 = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 4 = 0

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

x = \frac{- 4}{2 \cdot 1}

\frac{- 4}{2 \cdot 1} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

4 + x (x - 4) = x

x, 2 x^{2} + 3 y^{2} - x + 3 y + 6 = 0

(x - 1)^{2} - 4 = 44

2 x^{2} - 19 x + 45 = 0

- 3 y^{2} - 4 y

What are the solutions to the equation [-x-3]^2=5-2[-x] ?
The solutions to the equation [-x-3]^2=5-2[-x] are x=-2
Find the zeros of [-x-3]^2=5-2[-x]
The zeros of [-x-3]^2=5-2[-x] are x=-2