23-c(2c+3)=2(c+5)+c

Solution for 23-c(2c+3)=2(c+5)+c equation:

23-c(2c+3)=2(c+5)+c

We move all terms to the left:

23-c(2c+3)-(2(c+5)+c)=0

We multiply parentheses

-2c^2-3c-(2(c+5)+c)+23=0


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

2(c+5)+c

We add all the numbers together, and all the variables

c+2(c+5)

We multiply parentheses

c+2c+10

We add all the numbers together, and all the variables

3c+10

Back to the equation:

-(3c+10)


We get rid of parentheses

-2c^2-3c-3c-10+23=0

We add all the numbers together, and all the variables

-2c^2-6c+13=0

a = -2; b = -6; c = +13;
Δ = b2-4ac
Δ = -62-4·(-2)·13
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{35}}{2*-2}=\frac{6-2\sqrt{35}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{35}}{2*-2}=\frac{6+2\sqrt{35}}{-4}
$