60-c(3c+4)=4(c+6)+c

Solution for 60-c(3c+4)=4(c+6)+c equation:

60-c(3c+4)=4(c+6)+c

We move all terms to the left:

60-c(3c+4)-(4(c+6)+c)=0

We multiply parentheses

-3c^2-4c-(4(c+6)+c)+60=0


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

4(c+6)+c

We add all the numbers together, and all the variables

c+4(c+6)

We multiply parentheses

c+4c+24

We add all the numbers together, and all the variables

5c+24

Back to the equation:

-(5c+24)


We get rid of parentheses

-3c^2-4c-5c-24+60=0

We add all the numbers together, and all the variables

-3c^2-9c+36=0

a = -3; b = -9; c = +36;
Δ = b2-4ac
Δ = -92-4·(-3)·36
Δ = 513
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:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{57}}{2*-3}=\frac{9-3\sqrt{57}}{-6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{57}}{2*-3}=\frac{9+3\sqrt{57}}{-6}
$