What are the solutions to the equation 2x^2+4x-16=0 ?

The solutions to the equation 2x^2+4x-16=0 are x=2,x=-4

Find the zeros of 2x^2+4x-16=0

The zeros of 2x^2+4x-16=0 are x=2,x=-4

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2 x^{2} + 4 x - 16 = 0

x = 2, x = - 4

Solution steps

2 x^{2} + 4 x - 16 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 16) = 12

x_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 12}{2 \cdot 2}, x_{2} = \frac{- 4 - 12}{2 \cdot 2}

x = \frac{- 4 + 12}{2 \cdot 2} : 2

x = \frac{- 4 - 12}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

ln ∣ x + 2 ∣ = 0

\frac{a ^{- 4}}{a ^{- 2}}

1.01 \cdot 1 0^{- 4} = \frac{L - 3.9}{3.9}

- 3 x^{3} - 9 x^{2} + 54 x

- 2 < 2 + 4 x < 14

What are the solutions to the equation 2x^2+4x-16=0 ?
The solutions to the equation 2x^2+4x-16=0 are x=2,x=-4
Find the zeros of 2x^2+4x-16=0
The zeros of 2x^2+4x-16=0 are x=2,x=-4