What are the solutions to the equation 9=3m^2+2(m-13) ?

The solutions to the equation 9=3m^2+2(m-13) are m=(-1+sqrt(106))/3 ,m=-(1+sqrt(106))/3

Find the zeros of 9=3m^2+2(m-13)

The zeros of 9=3m^2+2(m-13) are m=(-1+sqrt(106))/3 ,m=-(1+sqrt(106))/3

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9 = 3 m^{2} + 2 (m - 13)

m = \frac{- 1 + 106}{3}, m = - \frac{1 + 106}{3}

Decimal Notation

m = 3.09854 \dots, m = - 3.76521 \dots

Solution steps

9 = 3 m^{2} + 2 (m - 13)

Expand

3 m^{2} + 2 (m - 13) : 3 m^{2} + 2 m - 26

9 = 3 m^{2} + 2 m - 26

Switch sides

3 m^{2} + 2 m - 26 = 9

Move

9

to the left side

3 m^{2} + 2 m - 35 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 ( - 35 )}{2 \cdot 3}

2^{2} - 4 \cdot 3 (- 35) = 2106

m_{1, 2} = \frac{- 2 \pm 2 106}{2 \cdot 3}

Separate the solutions

m_{1} = \frac{- 2 + 2 106}{2 \cdot 3}, m_{2} = \frac{- 2 - 2 106}{2 \cdot 3}

m = \frac{- 2 + 2 106}{2 \cdot 3} : \frac{- 1 + 106}{3}

m = \frac{- 2 - 2 106}{2 \cdot 3} : - \frac{1 + 106}{3}

The solutions to the quadratic equation are:

m = \frac{- 1 + 106}{3}, m = - \frac{1 + 106}{3}

- n^{2} = 3 - 4 n

v^{2} - 9. 8^{2} = 2 \cdot 9.8 \cdot 19.6

(4 x - 1)^{2} + (- 2)^{3} = 2020

(k + 2) \cdot x^{2} - 13 x + 4 k - 10 = 0

(7 - 2 f)^{2} = 9

What are the solutions to the equation 9=3m^2+2(m-13) ?
The solutions to the equation 9=3m^2+2(m-13) are m=(-1+sqrt(106))/3 ,m=-(1+sqrt(106))/3
Find the zeros of 9=3m^2+2(m-13)
The zeros of 9=3m^2+2(m-13) are m=(-1+sqrt(106))/3 ,m=-(1+sqrt(106))/3