What are the solutions to the equation x^2+12x=-2x-40 ?

The solutions to the equation x^2+12x=-2x-40 are x=-4,x=-10

Find the zeros of x^2+12x=-2x-40

The zeros of x^2+12x=-2x-40 are x=-4,x=-10

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

x = - 4, x = - 10

Solution steps

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

Move

40

to the left side

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

Move

2 x

to the left side

x^{2} + 14 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 40 = 6

x_{1, 2} = \frac{- 14 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 14 + 6}{2 \cdot 1}, x_{2} = \frac{- 14 - 6}{2 \cdot 1}

x = \frac{- 14 + 6}{2 \cdot 1} : - 4

x = \frac{- 14 - 6}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

x = - 4, x = - 10

k, - 5 k (x) x^{2} + 8 x = 4 (x + 2)

n^{2} - 7 n - 2 = 0

x^{2} + 10 x + 18

w^{2} + 16 w + 36 = 0

x^{2} - 5 x - 3 = 0

What are the solutions to the equation x^2+12x=-2x-40 ?
The solutions to the equation x^2+12x=-2x-40 are x=-4,x=-10
Find the zeros of x^2+12x=-2x-40
The zeros of x^2+12x=-2x-40 are x=-4,x=-10