(2/3)c+1=16-c

Solution for (2/3)c+1=16-c equation:

(2/3)c+1=16-c

We move all terms to the left:

(2/3)c+1-(16-c)=0


Domain of the equation: 3)c!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

(+2/3)c-(-1c+16)+1=0

We multiply parentheses

2c^2-(-1c+16)+1=0

We get rid of parentheses

2c^2+1c-16+1=0

We add all the numbers together, and all the variables

2c^2+c-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$