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

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

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

We move all terms to the left:

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

We multiply parentheses

-c^2+(6c-4)+3c-7=0

We get rid of parentheses

-c^2+6c+3c-4-7=0

We add all the numbers together, and all the variables

-1c^2+9c-11=0

a = -1; b = 9; c = -11;
Δ = b2-4ac
Δ = 92-4·(-1)·(-11)
Δ = 37
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{-(9)-\sqrt{37}}{2*-1}=\frac{-9-\sqrt{37}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{37}}{2*-1}=\frac{-9+\sqrt{37}}{-2}
$