/3c+15=-22c+17

Solution for /3c+15=-22c+17 equation:

/3c+15=-22c+17

We move all terms to the left:

/3c+15-(-22c+17)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We get rid of parentheses

/3c+22c-17+15=0

We multiply all the terms by the denominator


22c*3c-17*3c+15*3c+=0

We add all the numbers together, and all the variables

22c*3c-17*3c+15*3c=0

Wy multiply elements

66c^2-51c+45c=0

We add all the numbers together, and all the variables

66c^2-6c=0

a = 66; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·66·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*66}=\frac{0}{132}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*66}=\frac{12}{132}
=1/11
$