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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

-3c-(4c^2+24c)+56=0

We get rid of parentheses

-4c^2-3c-24c+56=0

We add all the numbers together, and all the variables

-4c^2-27c+56=0

a = -4; b = -27; c = +56;
Δ = b2-4ac
Δ = -272-4·(-4)·56
Δ = 1625
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:

$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-5\sqrt{65}}{2*-4}=\frac{27-5\sqrt{65}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+5\sqrt{65}}{2*-4}=\frac{27+5\sqrt{65}}{-8}
$