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

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

Find the zeros of 4a^2+16a+16=0

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

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4 a^{2} + 16 a + 16 = 0

a = - 2

Solution steps

4 a^{2} + 16 a + 16 = 0

Solve with the quadratic formula

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

1 6^{2} - 4 \cdot 4 \cdot 16 = 0

a_{1, 2} = \frac{- 16 \pm 0}{2 \cdot 4}

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

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

a = - 2

The solution to the quadratic equation is:

a = - 2

x, (x + 1)^{2} + y^{2} + (z + 4)^{2} = 5

x^{2} - 155 x + 5950 = 0

4 x = x^{2} - 21

3 x^{2} + 16 x - 75 = 0

x, f^{- 1} (x) = (x + 3)^{2}

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