What are the solutions to the equation x^2+4x+12=-2x+4 ?

The solutions to the equation x^2+4x+12=-2x+4 are x=-2,x=-4

Find the zeros of x^2+4x+12=-2x+4

The zeros of x^2+4x+12=-2x+4 are x=-2,x=-4

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x^{2} + 4 x + 12 = - 2 x + 4

x = - 2, x = - 4

Solution steps

x^{2} + 4 x + 12 = - 2 x + 4

Move

4

to the left side

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

Move

2 x

to the left side

x^{2} + 6 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 8 = 2

x_{1, 2} = \frac{- 6 \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 2}{2 \cdot 1}, x_{2} = \frac{- 6 - 2}{2 \cdot 1}

x = \frac{- 6 + 2}{2 \cdot 1} : - 2

x = \frac{- 6 - 2}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = - 2, x = - 4

x^{2} - 7 x + 6 = 0

4 x^{2} + 12 x + 14

- x^{2} + x + 6

0 = y^{2} - 65

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

What are the solutions to the equation x^2+4x+12=-2x+4 ?
The solutions to the equation x^2+4x+12=-2x+4 are x=-2,x=-4
Find the zeros of x^2+4x+12=-2x+4
The zeros of x^2+4x+12=-2x+4 are x=-2,x=-4