32-(2c+4)=2(c+4)c

Solution for 32-(2c+4)=2(c+4)c equation:

32-(2c+4)=2(c+4)c

We move all terms to the left:

32-(2c+4)-(2(c+4)c)=0

We get rid of parentheses

-2c-(2(c+4)c)-4+32=0


We calculate terms in parentheses: -(2(c+4)c), so:

2(c+4)c

We multiply parentheses

2c^2+8c

Back to the equation:

-(2c^2+8c)


We add all the numbers together, and all the variables

-2c-(2c^2+8c)+28=0

We get rid of parentheses

-2c^2-2c-8c+28=0

We add all the numbers together, and all the variables

-2c^2-10c+28=0

a = -2; b = -10; c = +28;
Δ = b2-4ac
Δ = -102-4·(-2)·28
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-18}{2*-2}=\frac{-8}{-4}
=+2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+18}{2*-2}=\frac{28}{-4}
=-7
$