What are the solutions to the equation 5m^2-4m-2=0 ?

The solutions to the equation 5m^2-4m-2=0 are m=(2+sqrt(14))/5 ,m=(2-sqrt(14))/5

Find the zeros of 5m^2-4m-2=0

The zeros of 5m^2-4m-2=0 are m=(2+sqrt(14))/5 ,m=(2-sqrt(14))/5

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

m = \frac{2 + 14}{5}, m = \frac{2 - 14}{5}

Decimal Notation

m = 1.14833 \dots, m = - 0.34833 \dots

Solution steps

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

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

(- 4)^{2} - 4 \cdot 5 (- 2) = 214

m_{1, 2} = \frac{- ( - 4 ) \pm 2 14}{2 \cdot 5}

Separate the solutions

m_{1} = \frac{- ( - 4 ) + 2 14}{2 \cdot 5}, m_{2} = \frac{- ( - 4 ) - 2 14}{2 \cdot 5}

m = \frac{- ( - 4 ) + 2 14}{2 \cdot 5} : \frac{2 + 14}{5}

m = \frac{- ( - 4 ) - 2 14}{2 \cdot 5} : \frac{2 - 14}{5}

The solutions to the quadratic equation are:

m = \frac{2 + 14}{5}, m = \frac{2 - 14}{5}

x (2 x - 4) = (2 x + 1) (x - 2)

25 x^{2} + 60 x = - 36

(- 2 - 4 i)^{2}

\frac{32 t}{32} = \frac{112}{32}

\frac{1}{3} \cdot 7

What are the solutions to the equation 5m^2-4m-2=0 ?
The solutions to the equation 5m^2-4m-2=0 are m=(2+sqrt(14))/5 ,m=(2-sqrt(14))/5
Find the zeros of 5m^2-4m-2=0
The zeros of 5m^2-4m-2=0 are m=(2+sqrt(14))/5 ,m=(2-sqrt(14))/5