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

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

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

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

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

m = 7, m = - 3

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} : 7

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

The solutions to the quadratic equation are:

m = 7, m = - 3

\frac{( 7 a ^{2} b ^{3} ) ^{2}}{a ^{3} b ^{5}}

x + 5 = 14 - \frac{x}{2}

2 x - (x - 1)^{2} = 3 x^{2} - 6

\frac{2 x - 3}{3 x - 5} > 3

3 x^{2} - 4 = 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=7,m=-3
Find the zeros of m^2-4m-21=0
The zeros of m^2-4m-21=0 are m=7,m=-3