What are the solutions to the equation 2x^2+8x=x^2-16 ?

The solutions to the equation 2x^2+8x=x^2-16 are x=-4

Find the zeros of 2x^2+8x=x^2-16

The zeros of 2x^2+8x=x^2-16 are x=-4

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2 x^{2} + 8 x = x^{2} - 16

x = - 4

Solution steps

2 x^{2} + 8 x = x^{2} - 16

Move

16

to the left side

2 x^{2} + 8 x + 16 = x^{2}

Move

x^{2}

to the left side

x^{2} + 8 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 16 = 0

x_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 1}

x = \frac{- 8}{2 \cdot 1}

\frac{- 8}{2 \cdot 1} = - 4

x = - 4

The solution to the quadratic equation is:

x = - 4

x^{2} + 12 x + 36

x^{2} - 6 x - 16 = 0

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

x^{2} - 2 x + 1 = 3

x, x z + y z^{3} = 2 + x^{2} e^{y}

What are the solutions to the equation 2x^2+8x=x^2-16 ?
The solutions to the equation 2x^2+8x=x^2-16 are x=-4
Find the zeros of 2x^2+8x=x^2-16
The zeros of 2x^2+8x=x^2-16 are x=-4