13-(2c+2)=2(c2+2)+3c

Solution for 13-(2c+2)=2(c2+2)+3c equation:

13-(2c+2)=2(c2+2)+3c

We move all terms to the left:

13-(2c+2)-(2(c2+2)+3c)=0

We add all the numbers together, and all the variables

-(2(+c^2+2)+3c)-(2c+2)+13=0

We get rid of parentheses

-(2(+c^2+2)+3c)-2c-2+13=0


We calculate terms in parentheses: -(2(+c^2+2)+3c), so:

2(+c^2+2)+3c

We multiply parentheses

2c^2+3c+4

Back to the equation:

-(2c^2+3c+4)


We add all the numbers together, and all the variables

-2c-(2c^2+3c+4)+11=0

We get rid of parentheses

-2c^2-2c-3c-4+11=0

We add all the numbers together, and all the variables

-2c^2-5c+7=0

a = -2; b = -5; c = +7;
Δ = b2-4ac
Δ = -52-4·(-2)·7
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*-2}=\frac{-4}{-4}
=1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*-2}=\frac{14}{-4}
=-3+1/2
$