What are the solutions to the equation m^2+20m-4=0 ?

The solutions to the equation m^2+20m-4=0 are m=2(sqrt(26)-5),m=-2(5+sqrt(26))

Find the zeros of m^2+20m-4=0

The zeros of m^2+20m-4=0 are m=2(sqrt(26)-5),m=-2(5+sqrt(26))

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m^{2} + 20 m - 4 = 0

m = 2 (26 - 5), m = - 2 (5 + 26)

Decimal Notation

m = 0.19803 \dots, m = - 20.19803 \dots

Solution steps

m^{2} + 20 m - 4 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 4) = 426

m_{1, 2} = \frac{- 20 \pm 4 26}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 20 + 4 26}{2 \cdot 1}, m_{2} = \frac{- 20 - 4 26}{2 \cdot 1}

m = \frac{- 20 + 4 26}{2 \cdot 1} : 2 (26 - 5)

m = \frac{- 20 - 4 26}{2 \cdot 1} : - 2 (5 + 26)

The solutions to the quadratic equation are:

m = 2 (26 - 5), m = - 2 (5 + 26)

54 x^{3} - 128

1 = x^{2} + 2 x \cdot 0.2 + 0.04

(4 x + 1) (x - 8) = 0

\frac{5 !}{2 ! ( 5 - 2 ) !}

\frac{16 a ^{4} - 40 a ^{2} + 24 a}{12 a ^{3}}

What are the solutions to the equation m^2+20m-4=0 ?
The solutions to the equation m^2+20m-4=0 are m=2(sqrt(26)-5),m=-2(5+sqrt(26))
Find the zeros of m^2+20m-4=0
The zeros of m^2+20m-4=0 are m=2(sqrt(26)-5),m=-2(5+sqrt(26))