What are the solutions to the equation (m+2)(3m+10)=3(m+5)+15 ?

The solutions to the equation (m+2)(3m+10)=3(m+5)+15 are m= 2/3 ,m=-5

Find the zeros of (m+2)(3m+10)=3(m+5)+15

The zeros of (m+2)(3m+10)=3(m+5)+15 are m= 2/3 ,m=-5

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(m + 2) (3 m + 10) = 3 (m + 5) + 15

m = \frac{2}{3}, m = - 5

Solution steps

(m + 2) (3 m + 10) = 3 (m + 5) + 15

Expand

(m + 2) (3 m + 10) : 3 m^{2} + 16 m + 20

Expand

3 (m + 5) + 15 : 3 m + 30

3 m^{2} + 16 m + 20 = 3 m + 30

Move

30

to the left side

3 m^{2} + 16 m - 10 = 3 m

Move

3 m

to the left side

3 m^{2} + 13 m - 10 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 3 ( - 10 )}{2 \cdot 3}

1 3^{2} - 4 \cdot 3 (- 10) = 17

m_{1, 2} = \frac{- 13 \pm 17}{2 \cdot 3}

Separate the solutions

m_{1} = \frac{- 13 + 17}{2 \cdot 3}, m_{2} = \frac{- 13 - 17}{2 \cdot 3}

m = \frac{- 13 + 17}{2 \cdot 3} : \frac{2}{3}

m = \frac{- 13 - 17}{2 \cdot 3} : - 5

The solutions to the quadratic equation are:

m = \frac{2}{3}, m = - 5

\frac{1}{4} (1 + 3 (- 0.33))

4 - 6 x = x - 24

- \frac{2}{3} (\frac{6}{5} x - \frac{7}{10}) = \frac{17}{20}

(6 x^{5} y^{3})^{2}

5 x + 20

What are the solutions to the equation (m+2)(3m+10)=3(m+5)+15 ?
The solutions to the equation (m+2)(3m+10)=3(m+5)+15 are m= 2/3 ,m=-5
Find the zeros of (m+2)(3m+10)=3(m+5)+15
The zeros of (m+2)(3m+10)=3(m+5)+15 are m= 2/3 ,m=-5