What are the solutions to the equation m^2+12m+19=0 ?

The solutions to the equation m^2+12m+19=0 are m=-6+sqrt(17),m=-6-sqrt(17)

Find the zeros of m^2+12m+19=0

The zeros of m^2+12m+19=0 are m=-6+sqrt(17),m=-6-sqrt(17)

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m^{2} + 12 m + 19 = 0

m = - 6 + 17, m = - 6 - 17

Decimal Notation

m = - 1.87689 \dots, m = - 10.12310 \dots

Solution steps

m^{2} + 12 m + 19 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 19 = 217

m_{1, 2} = \frac{- 12 \pm 2 17}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 12 + 2 17}{2 \cdot 1}, m_{2} = \frac{- 12 - 2 17}{2 \cdot 1}

m = \frac{- 12 + 2 17}{2 \cdot 1} : - 6 + 17

m = \frac{- 12 - 2 17}{2 \cdot 1} : - 6 - 17

The solutions to the quadratic equation are:

m = - 6 + 17, m = - 6 - 17

x^{2} - 2 x - 48 = 0

- 6 x - 7 = 23

0 = x^{2} - 2

3 - (- 4)

a^{2} b^{2} - (ab - 7)^{2}

What are the solutions to the equation m^2+12m+19=0 ?
The solutions to the equation m^2+12m+19=0 are m=-6+sqrt(17),m=-6-sqrt(17)
Find the zeros of m^2+12m+19=0
The zeros of m^2+12m+19=0 are m=-6+sqrt(17),m=-6-sqrt(17)