2c+6+3c+4+(1/2)c+5=180

Solution for 2c+6+3c+4+(1/2)c+5=180 equation:

2c+6+3c+4+(1/2)c+5=180

We move all terms to the left:

2c+6+3c+4+(1/2)c+5-(180)=0


Domain of the equation: 2)c!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

2c+3c+(+1/2)c+6+4+5-180=0

We add all the numbers together, and all the variables

5c+(+1/2)c-165=0

We multiply parentheses

c^2+5c-165=0

a = 1; b = 5; c = -165;
Δ = b2-4ac
Δ = 52-4·1·(-165)
Δ = 685
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{685}}{2*1}=\frac{-5-\sqrt{685}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{685}}{2*1}=\frac{-5+\sqrt{685}}{2}
$