What are the solutions to the equation m^2+4m-21=0 ?

The solutions to the equation m^2+4m-21=0 are m=3,m=-7

Find the zeros of m^2+4m-21=0

The zeros of m^2+4m-21=0 are m=3,m=-7

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m^{2} + 4 m - 21 = 0

m = 3, m = - 7

Solution steps

m^{2} + 4 m - 21 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 21) = 10

m_{1, 2} = \frac{- 4 \pm 10}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 4 + 10}{2 \cdot 1}, m_{2} = \frac{- 4 - 10}{2 \cdot 1}

m = \frac{- 4 + 10}{2 \cdot 1} : 3

m = \frac{- 4 - 10}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

m = 3, m = - 7

4 (x - 2)^{2} = 36

3 x (2 x + 5) - 6 (3 - x) (x - 1) = 14 x (x - 1)

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

- 4 x^{2} + 2 x - 3 = 0

y, x^{2} - 4 x y + 4 y^{2} + x - 12 y - 10 = 0

What are the solutions to the equation m^2+4m-21=0 ?
The solutions to the equation m^2+4m-21=0 are m=3,m=-7
Find the zeros of m^2+4m-21=0
The zeros of m^2+4m-21=0 are m=3,m=-7