c=(2c+3)(6c-5)

Solution for c=(2c+3)(6c-5) equation:

c=(2c+3)(6c-5)

We move all terms to the left:

c-((2c+3)(6c-5))=0

We multiply parentheses ..

-((+12c^2-10c+18c-15))+c=0


We calculate terms in parentheses: -((+12c^2-10c+18c-15)), so:

(+12c^2-10c+18c-15)

We get rid of parentheses

12c^2-10c+18c-15

We add all the numbers together, and all the variables

12c^2+8c-15

Back to the equation:

-(12c^2+8c-15)


We add all the numbers together, and all the variables

c-(12c^2+8c-15)=0

We get rid of parentheses

-12c^2+c-8c+15=0

We add all the numbers together, and all the variables

-12c^2-7c+15=0

a = -12; b = -7; c = +15;
Δ = b2-4ac
Δ = -72-4·(-12)·15
Δ = 769
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{-(-7)-\sqrt{769}}{2*-12}=\frac{7-\sqrt{769}}{-24}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{769}}{2*-12}=\frac{7+\sqrt{769}}{-24}
$