2/5c-4/5c-6=c+7

Solution for 2/5c-4/5c-6=c+7 equation:

2/5c-4/5c-6=c+7

We move all terms to the left:

2/5c-4/5c-6-(c+7)=0


Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R


We get rid of parentheses

2/5c-4/5c-c-7-6=0

We multiply all the terms by the denominator


-c*5c-7*5c-6*5c+2-4=0

We add all the numbers together, and all the variables

-c*5c-7*5c-6*5c-2=0

Wy multiply elements

-5c^2-35c-30c-2=0

We add all the numbers together, and all the variables

-5c^2-65c-2=0

a = -5; b = -65; c = -2;
Δ = b2-4ac
Δ = -652-4·(-5)·(-2)
Δ = 4185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4185}=\sqrt{9*465}=\sqrt{9}*\sqrt{465}=3\sqrt{465}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-3\sqrt{465}}{2*-5}=\frac{65-3\sqrt{465}}{-10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+3\sqrt{465}}{2*-5}=\frac{65+3\sqrt{465}}{-10}
$