What are the solutions to the equation m^2-m=1 ?

The solutions to the equation m^2-m=1 are m=(1+sqrt(5))/2 ,m=(1-sqrt(5))/2

Find the zeros of m^2-m=1

The zeros of m^2-m=1 are m=(1+sqrt(5))/2 ,m=(1-sqrt(5))/2

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m^{2} - m = 1

m = \frac{1 + 5}{2}, m = \frac{1 - 5}{2}

Decimal Notation

m = 1.61803 \dots, m = - 0.61803 \dots

Solution steps

m^{2} - m = 1

Move

1

to the left side

m^{2} - m - 1 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 1) = 5

m_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, m_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

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

m = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

The solutions to the quadratic equation are:

m = \frac{1 + 5}{2}, m = \frac{1 - 5}{2}

0 = x^{2} - \frac{3}{2} x - \frac{70}{8}

x, x^{2} y + 3 y^{3} = 24

6 x^{2} - 6 x - 12

x^{2} + 40 = - 14 x

y, x - \frac{x ^{3}}{4} - \frac{x y ^{2}}{3} = 0

What are the solutions to the equation m^2-m=1 ?
The solutions to the equation m^2-m=1 are m=(1+sqrt(5))/2 ,m=(1-sqrt(5))/2
Find the zeros of m^2-m=1
The zeros of m^2-m=1 are m=(1+sqrt(5))/2 ,m=(1-sqrt(5))/2