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

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

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

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

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

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

Decimal Notation

y = - 0.34314 \dots, y = - 11.65685 \dots

Solution steps

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

Solve with the quadratic formula

y_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(2) x + 9 = 2 x - 3

\frac{27}{49}

- 4 < \frac{2 x - 3}{3} < 4

(\frac{3000}{32}) r^{2} + 7.13 r + 26000 = 0

\frac{14}{2} - 5

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