What are the solutions to the equation 3m(m-1)-9=2m(m-3)+4 ?

The solutions to the equation 3m(m-1)-9=2m(m-3)+4 are m=(-3+sqrt(61))/2 ,m=(-3-sqrt(61))/2

Find the zeros of 3m(m-1)-9=2m(m-3)+4

The zeros of 3m(m-1)-9=2m(m-3)+4 are m=(-3+sqrt(61))/2 ,m=(-3-sqrt(61))/2

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3 m (m - 1) - 9 = 2 m (m - 3) + 4

m = \frac{- 3 + 61}{2}, m = \frac{- 3 - 61}{2}

Decimal Notation

m = 2.40512 \dots, m = - 5.40512 \dots

Solution steps

3 m (m - 1) - 9 = 2 m (m - 3) + 4

Expand

3 m (m - 1) - 9 : 3 m^{2} - 3 m - 9

Expand

2 m (m - 3) + 4 : 2 m^{2} - 6 m + 4

3 m^{2} - 3 m - 9 = 2 m^{2} - 6 m + 4

Move

4

to the left side

3 m^{2} - 3 m - 13 = 2 m^{2} - 6 m

Move

6 m

to the left side

3 m^{2} + 3 m - 13 = 2 m^{2}

Move

2 m^{2}

to the left side

m^{2} + 3 m - 13 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 13) = 61

m_{1, 2} = \frac{- 3 \pm 61}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 3 + 61}{2 \cdot 1}, m_{2} = \frac{- 3 - 61}{2 \cdot 1}

m = \frac{- 3 + 61}{2 \cdot 1} : \frac{- 3 + 61}{2}

m = \frac{- 3 - 61}{2 \cdot 1} : \frac{- 3 - 61}{2}

The solutions to the quadratic equation are:

m = \frac{- 3 + 61}{2}, m = \frac{- 3 - 61}{2}

4 w^{2} + 12 w = - 9

- 0.2 x^{2} + 3 x + 0.25 = 0

- 110 = 10 - 5 x^{2}

x^{2} + 13 x + 27 = 0

- 2^{2} + 3^{2} = c^{2}

What are the solutions to the equation 3m(m-1)-9=2m(m-3)+4 ?
The solutions to the equation 3m(m-1)-9=2m(m-3)+4 are m=(-3+sqrt(61))/2 ,m=(-3-sqrt(61))/2
Find the zeros of 3m(m-1)-9=2m(m-3)+4
The zeros of 3m(m-1)-9=2m(m-3)+4 are m=(-3+sqrt(61))/2 ,m=(-3-sqrt(61))/2