42-(2c+4)=4(c+6)c

Solution for 42-(2c+4)=4(c+6)c equation:

42-(2c+4)=4(c+6)c

We move all terms to the left:

42-(2c+4)-(4(c+6)c)=0

We get rid of parentheses

-2c-(4(c+6)c)-4+42=0


We calculate terms in parentheses: -(4(c+6)c), so:

4(c+6)c

We multiply parentheses

4c^2+24c

Back to the equation:

-(4c^2+24c)


We add all the numbers together, and all the variables

-2c-(4c^2+24c)+38=0

We get rid of parentheses

-4c^2-2c-24c+38=0

We add all the numbers together, and all the variables

-4c^2-26c+38=0

a = -4; b = -26; c = +38;
Δ = b2-4ac
Δ = -262-4·(-4)·38
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{321}}{2*-4}=\frac{26-2\sqrt{321}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{321}}{2*-4}=\frac{26+2\sqrt{321}}{-8}
$