2c-3=2(6-c)7c

Solution for 2c-3=2(6-c)7c equation:

2c-3=2(6-c)7c

We move all terms to the left:

2c-3-(2(6-c)7c)=0

We add all the numbers together, and all the variables

2c-(2(-1c+6)7c)-3=0


We calculate terms in parentheses: -(2(-1c+6)7c), so:

2(-1c+6)7c

We multiply parentheses

-14c^2+84c

Back to the equation:

-(-14c^2+84c)


We get rid of parentheses

14c^2-84c+2c-3=0

We add all the numbers together, and all the variables

14c^2-82c-3=0

a = 14; b = -82; c = -3;
Δ = b2-4ac
Δ = -822-4·14·(-3)
Δ = 6892
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:

$\sqrt{\Delta}=\sqrt{6892}=\sqrt{4*1723}=\sqrt{4}*\sqrt{1723}=2\sqrt{1723}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-82)-2\sqrt{1723}}{2*14}=\frac{82-2\sqrt{1723}}{28}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-82)+2\sqrt{1723}}{2*14}=\frac{82+2\sqrt{1723}}{28}
$