What are the solutions to the equation x^2+12x+29=0 ?

The solutions to the equation x^2+12x+29=0 are x=-6+sqrt(7),x=-6-sqrt(7)

Find the zeros of x^2+12x+29=0

The zeros of x^2+12x+29=0 are x=-6+sqrt(7),x=-6-sqrt(7)

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x^{2} + 12 x + 29 = 0

x = - 6 + 7, x = - 6 - 7

Solution steps

x^{2} + 12 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 29 = 27

x_{1, 2} = \frac{- 12 \pm 2 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 2 7}{2 \cdot 1}, x_{2} = \frac{- 12 - 2 7}{2 \cdot 1}

x = \frac{- 12 + 2 7}{2 \cdot 1} : - 6 + 7

x = \frac{- 12 - 2 7}{2 \cdot 1} : - 6 - 7

The solutions to the quadratic equation are:

x = - 6 + 7, x = - 6 - 7

\frac{6 ^{- 2}}{( - 1 ) ^{- 4}}

2 - 10 x = (3 x - 7) (x + 1)

- 3 x - (- 5) \leq 20

\frac{2 + x}{7} = x + 6

- 9 d^{4} + d^{3} - 2 d^{2}

What are the solutions to the equation x^2+12x+29=0 ?
The solutions to the equation x^2+12x+29=0 are x=-6+sqrt(7),x=-6-sqrt(7)
Find the zeros of x^2+12x+29=0
The zeros of x^2+12x+29=0 are x=-6+sqrt(7),x=-6-sqrt(7)