2(5+4c)+15=-2c-40

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Solution for 2(5+4c)+15=-2c-40 equation: 



2(5+4c)+15=-2c-40

We move all terms to the left:

2(5+4c)+15-(-2c-40)=0

We add all the numbers together, and all the variables

2(4c+5)-(-2c-40)+15=0

We multiply parentheses

8c-(-2c-40)+10+15=0

We get rid of parentheses

8c+2c+40+10+15=0

We add all the numbers together, and all the variables

10c+65=0

We move all terms containing c to the left, all other terms to the right

10c=-65

c=-65/10

c=-6+1/2

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