What are the solutions to the equation (2)2m^2-5m=0, 5/2 ,m=2 ?

The solutions to the equation (2)2m^2-5m=0, 5/2 ,m=2 are No Solution

Find the zeros of (2)2m^2-5m=0, 5/2 ,m=2

The zeros of (2)2m^2-5m=0, 5/2 ,m=2 are No Solution

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(2) 2 m^{2} - 5 m = 0, \frac{5}{2}, m = 2

No Solution

Solution steps

(2) \cdot 2 m^{2} - 5 m = 0

Expand

(2) \cdot 2 m^{2} - 5 m : 4 m^{2} - 5 m

4 m^{2} - 5 m = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 5)^{2} - 4 \cdot 4 \cdot 0 = 5

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

Separate the solutions

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

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

m = \frac{- ( - 5 ) - 5}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

m = \frac{5}{4}, m = 0

Check constrains

No Solution

- 1.25 x^{2} + 35 x - 175 = 50

\frac{( 6 x ^{2} )}{100} + \frac{( 6 x )}{10} + 5 = 6

11 p + 4 = 3 p^{2}

3 x^{2} + 7 x + 4 x = 0

t, e = v_{c} \cdot t - \frac{1}{2} \cdot a \cdot t^{2}

What are the solutions to the equation (2)2m^2-5m=0, 5/2 ,m=2 ?
The solutions to the equation (2)2m^2-5m=0, 5/2 ,m=2 are No Solution
Find the zeros of (2)2m^2-5m=0, 5/2 ,m=2
The zeros of (2)2m^2-5m=0, 5/2 ,m=2 are No Solution