(4c^2-3c-2)+(3c^2-4c-2)=c

Solution for (4c^2-3c-2)+(3c^2-4c-2)=c equation:

(4c^2-3c-2)+(3c^2-4c-2)=c

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1c+(4c^2-3c-2)+(3c^2-4c-2)=0

We get rid of parentheses

4c^2+3c^2-1c-3c-4c-2-2=0

We add all the numbers together, and all the variables

7c^2-8c-4=0

a = 7; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·7·(-4)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{11}}{2*7}=\frac{8-4\sqrt{11}}{14}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{11}}{2*7}=\frac{8+4\sqrt{11}}{14}
$