27-(c2+4)=2(c+4)+c

Solution for 27-(c2+4)=2(c+4)+c equation:

27-(c2+4)=2(c+4)+c

We move all terms to the left:

27-(c2+4)-(2(c+4)+c)=0

We add all the numbers together, and all the variables

-(+c^2+4)-(2(c+4)+c)+27=0

We get rid of parentheses

-c^2-(2(c+4)+c)-4+27=0


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

2(c+4)+c

We add all the numbers together, and all the variables

c+2(c+4)

We multiply parentheses

c+2c+8

We add all the numbers together, and all the variables

3c+8

Back to the equation:

-(3c+8)


We add all the numbers together, and all the variables

-1c^2-(3c+8)+23=0

We get rid of parentheses

-1c^2-3c-8+23=0

We add all the numbers together, and all the variables

-1c^2-3c+15=0

a = -1; b = -3; c = +15;
Δ = b2-4ac
Δ = -32-4·(-1)·15
Δ = 69
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{69}}{2*-1}=\frac{3-\sqrt{69}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{69}}{2*-1}=\frac{3+\sqrt{69}}{-2}
$