What are the solutions to the equation m^2-4m-32=0 ?

The solutions to the equation m^2-4m-32=0 are m=8,m=-4

Find the zeros of m^2-4m-32=0

The zeros of m^2-4m-32=0 are m=8,m=-4

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m^{2} - 4 m - 32 = 0

m = 8, m = - 4

Solution steps

m^{2} - 4 m - 32 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 32) = 12

m_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 1}, m_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 1}

m = \frac{- ( - 4 ) + 12}{2 \cdot 1} : 8

m = \frac{- ( - 4 ) - 12}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

m = 8, m = - 4

A, (A + X)^{2} = BC

x^{2} - x = 0

2 x^{2} + 3 x - 5 = 0

m^{2} + 96 m + \frac{8}{3} = 0

8 x^{2} = 4

What are the solutions to the equation m^2-4m-32=0 ?
The solutions to the equation m^2-4m-32=0 are m=8,m=-4
Find the zeros of m^2-4m-32=0
The zeros of m^2-4m-32=0 are m=8,m=-4