What are the solutions to the equation a^2+12a+4=0 ?

The solutions to the equation a^2+12a+4=0 are a=2(2sqrt(2)-3),a=-2(3+2sqrt(2))

Find the zeros of a^2+12a+4=0

The zeros of a^2+12a+4=0 are a=2(2sqrt(2)-3),a=-2(3+2sqrt(2))

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a^{2} + 12 a + 4 = 0

a = 2 (22 - 3), a = - 2 (3 + 22)

Decimal Notation

a = - 0.34314 \dots, a = - 11.65685 \dots

Solution steps

a^{2} + 12 a + 4 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 1 \cdot 4 = 82

a_{1, 2} = \frac{- 12 \pm 8 2}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 12 + 8 2}{2 \cdot 1}, a_{2} = \frac{- 12 - 8 2}{2 \cdot 1}

a = \frac{- 12 + 8 2}{2 \cdot 1} : 2 (22 - 3)

a = \frac{- 12 - 8 2}{2 \cdot 1} : - 2 (3 + 22)

The solutions to the quadratic equation are:

a = 2 (22 - 3), a = - 2 (3 + 22)

5 x^{2} - 10 x - 40 = 0

5 x (x - 2021) + x - 2021 = 0

x = 10 (4.7 - x)^{2}

x, x^{2} + 4 x y + 16 x^{2} = 0

(3 x - 2)^{2} = 121

What are the solutions to the equation a^2+12a+4=0 ?
The solutions to the equation a^2+12a+4=0 are a=2(2sqrt(2)-3),a=-2(3+2sqrt(2))
Find the zeros of a^2+12a+4=0
The zeros of a^2+12a+4=0 are a=2(2sqrt(2)-3),a=-2(3+2sqrt(2))