What are the solutions to the equation m^2+80m+1200=0 ?

The solutions to the equation m^2+80m+1200=0 are m=-20,m=-60

Find the zeros of m^2+80m+1200=0

The zeros of m^2+80m+1200=0 are m=-20,m=-60

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m^{2} + 80 m + 1200 = 0

m = - 20, m = - 60

Solution steps

m^{2} + 80 m + 1200 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 1 \cdot 1200}{2 \cdot 1}

8 0^{2} - 4 \cdot 1 \cdot 1200 = 40

m_{1, 2} = \frac{- 80 \pm 40}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 80 + 40}{2 \cdot 1}, m_{2} = \frac{- 80 - 40}{2 \cdot 1}

m = \frac{- 80 + 40}{2 \cdot 1} : - 20

m = \frac{- 80 - 40}{2 \cdot 1} : - 60

The solutions to the quadratic equation are:

m = - 20, m = - 60

x, x^{2} - 6 x + 13 = 0

\frac{y ( 2 y - 1 )}{2} = \frac{1 - y}{3}

4 x^{2} + 8 x + 3 = 8

8 - 2 x - x^{2} = 0

20 x^{2} - 30 x = 2 x + 45

What are the solutions to the equation m^2+80m+1200=0 ?
The solutions to the equation m^2+80m+1200=0 are m=-20,m=-60
Find the zeros of m^2+80m+1200=0
The zeros of m^2+80m+1200=0 are m=-20,m=-60