What are the solutions to the equation x^2+4x+1=7-x^2 ?

The solutions to the equation x^2+4x+1=7-x^2 are x=1,x=-3

Find the zeros of x^2+4x+1=7-x^2

The zeros of x^2+4x+1=7-x^2 are x=1,x=-3

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x^{2} + 4 x + 1 = 7 - x^{2}

x = 1, x = - 3

Solution steps

x^{2} + 4 x + 1 = 7 - x^{2}

Move

x^{2}

to the left side

2 x^{2} + 4 x + 1 = 7

Move

7

to the left side

2 x^{2} + 4 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 6) = 8

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = - 3

2 x^{2} + 3 x - 4 = 0

- 3 x^{2} = 72

2 x^{2} + 4 x - 3 = 0

4.9 t^{2} - 25 t - 50 = 0

x^{2} + (8) x + (8) + 8 = 0

What are the solutions to the equation x^2+4x+1=7-x^2 ?
The solutions to the equation x^2+4x+1=7-x^2 are x=1,x=-3
Find the zeros of x^2+4x+1=7-x^2
The zeros of x^2+4x+1=7-x^2 are x=1,x=-3