(6c-8)(5=3c)=14

Solution for (6c-8)(5=3c)=14 equation:

(6c-8)(5=3c)=14

We move all terms to the left:

(6c-8)(5-(3c))=0

We add all the numbers together, and all the variables

(6c-8)(-3c+5)=0

We multiply parentheses ..

(-18c^2+30c+24c-40)=0

We get rid of parentheses

-18c^2+30c+24c-40=0

We add all the numbers together, and all the variables

-18c^2+54c-40=0

a = -18; b = 54; c = -40;
Δ = b2-4ac
Δ = 542-4·(-18)·(-40)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-6}{2*-18}=\frac{-60}{-36}
=1+2/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+6}{2*-18}=\frac{-48}{-36}
=1+1/3
$