2c-(c-5)=c(2c-5)-5

Solution for 2c-(c-5)=c(2c-5)-5 equation:

2c-(c-5)=c(2c-5)-5

We move all terms to the left:

2c-(c-5)-(c(2c-5)-5)=0

We get rid of parentheses

2c-c-(c(2c-5)-5)+5=0


We calculate terms in parentheses: -(c(2c-5)-5), so:

c(2c-5)-5

We multiply parentheses

2c^2-5c-5

Back to the equation:

-(2c^2-5c-5)


We add all the numbers together, and all the variables

c-(2c^2-5c-5)+5=0

We get rid of parentheses

-2c^2+c+5c+5+5=0

We add all the numbers together, and all the variables

-2c^2+6c+10=0

a = -2; b = 6; c = +10;
Δ = b2-4ac
Δ = 62-4·(-2)·10
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{29}}{2*-2}=\frac{-6-2\sqrt{29}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{29}}{2*-2}=\frac{-6+2\sqrt{29}}{-4}
$