What are the solutions to the equation 2x^2+12x+16=0 ?

The solutions to the equation 2x^2+12x+16=0 are x=-2,x=-4

Find the zeros of 2x^2+12x+16=0

The zeros of 2x^2+12x+16=0 are x=-2,x=-4

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2 x^{2} + 12 x + 16 = 0

x = - 2, x = - 4

Solution steps

2 x^{2} + 12 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 16}{2 \cdot 2}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2, x = - 4

- 2 m < 2 and 11 m > 22

\frac{15}{16}

\frac{x ^{3} - y ^{3}}{x + y} \cdot \frac{x ^{2} - y ^{2}}{x ^{2} + x y + y ^{2}}

a - 6 \geq 13

62 + 12

What are the solutions to the equation 2x^2+12x+16=0 ?
The solutions to the equation 2x^2+12x+16=0 are x=-2,x=-4
Find the zeros of 2x^2+12x+16=0
The zeros of 2x^2+12x+16=0 are x=-2,x=-4