c=(3c-4)(6c+7)

Solution for c=(3c-4)(6c+7) equation:

c=(3c-4)(6c+7)

We move all terms to the left:

c-((3c-4)(6c+7))=0

We multiply parentheses ..

-((+18c^2+21c-24c-28))+c=0


We calculate terms in parentheses: -((+18c^2+21c-24c-28)), so:

(+18c^2+21c-24c-28)

We get rid of parentheses

18c^2+21c-24c-28

We add all the numbers together, and all the variables

18c^2-3c-28

Back to the equation:

-(18c^2-3c-28)


We add all the numbers together, and all the variables

c-(18c^2-3c-28)=0

We get rid of parentheses

-18c^2+c+3c+28=0

We add all the numbers together, and all the variables

-18c^2+4c+28=0

a = -18; b = 4; c = +28;
Δ = b2-4ac
Δ = 42-4·(-18)·28
Δ = 2032
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:

$\sqrt{\Delta}=\sqrt{2032}=\sqrt{16*127}=\sqrt{16}*\sqrt{127}=4\sqrt{127}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{127}}{2*-18}=\frac{-4-4\sqrt{127}}{-36}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{127}}{2*-18}=\frac{-4+4\sqrt{127}}{-36}
$