What are the solutions to the equation 2x^2+16x+32=0 ?

The solutions to the equation 2x^2+16x+32=0 are x=-4

Find the zeros of 2x^2+16x+32=0

The zeros of 2x^2+16x+32=0 are x=-4

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2 x^{2} + 16 x + 32 = 0

x = - 4

Solution steps

2 x^{2} + 16 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 \cdot 32 = 0

x_{1, 2} = \frac{- 16 \pm 0}{2 \cdot 2}

x = \frac{- 16}{2 \cdot 2}

\frac{- 16}{2 \cdot 2} = - 4

x = - 4

The solution to the quadratic equation is:

x = - 4

(x + 2)^{2} + (x - 3)^{2} = 9

- \frac{1}{6} x^{2} + x - 3 = 0

2^{2} + 2^{2} = x^{2}

a^{2} - 1 = 0

3 x^{2} - 1 = 47

What are the solutions to the equation 2x^2+16x+32=0 ?
The solutions to the equation 2x^2+16x+32=0 are x=-4
Find the zeros of 2x^2+16x+32=0
The zeros of 2x^2+16x+32=0 are x=-4