What are the solutions to the equation m^2-4m-24=0 ?

The solutions to the equation m^2-4m-24=0 are m=2(1+sqrt(7)),m=2(1-sqrt(7))

Find the zeros of m^2-4m-24=0

The zeros of m^2-4m-24=0 are m=2(1+sqrt(7)),m=2(1-sqrt(7))

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m^{2} - 4 m - 24 = 0

m = 2 (1 + 7), m = 2 (1 - 7)

Decimal Notation

m = 7.29150 \dots, m = - 3.29150 \dots

Solution steps

m^{2} - 4 m - 24 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 24) = 47

m_{1, 2} = \frac{- ( - 4 ) \pm 4 7}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 4 ) + 4 7}{2 \cdot 1}, m_{2} = \frac{- ( - 4 ) - 4 7}{2 \cdot 1}

m = \frac{- ( - 4 ) + 4 7}{2 \cdot 1} : 2 (1 + 7)

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

The solutions to the quadratic equation are:

m = 2 (1 + 7), m = 2 (1 - 7)

3 u^{2} - 10 u + 3 = 0

4 x^{2} + 32 x + 28 = 0

4 y^{2} - 6 y + 9 = 0

y, x = a y^{2} + b y + c

11 x^{2} - 7 x + 1 = 0

What are the solutions to the equation m^2-4m-24=0 ?
The solutions to the equation m^2-4m-24=0 are m=2(1+sqrt(7)),m=2(1-sqrt(7))
Find the zeros of m^2-4m-24=0
The zeros of m^2-4m-24=0 are m=2(1+sqrt(7)),m=2(1-sqrt(7))