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

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

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

We move all terms to the left:

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

We get rid of parentheses

-2c-(2(c+4)c)-4+27=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)+23=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = -10; c = +23;
Δ = b2-4ac
Δ = -102-4·(-2)·23
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{71}}{2*-2}=\frac{10-2\sqrt{71}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{71}}{2*-2}=\frac{10+2\sqrt{71}}{-4}
$