(6c-8)(5+3c)=14^2

Solution for (6c-8)(5+3c)=14^2 equation:

(6c-8)(5+3c)=14^2

We move all terms to the left:

(6c-8)(5+3c)-(14^2)=0

We add all the numbers together, and all the variables

(6c-8)(3c+5)-14^2=0

We add all the numbers together, and all the variables

(6c-8)(3c+5)-196=0

We multiply parentheses ..

(+18c^2+30c-24c-40)-196=0

We get rid of parentheses

18c^2+30c-24c-40-196=0

We add all the numbers together, and all the variables

18c^2+6c-236=0

a = 18; b = 6; c = -236;
Δ = b2-4ac
Δ = 62-4·18·(-236)
Δ = 17028
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:

$\sqrt{\Delta}=\sqrt{17028}=\sqrt{36*473}=\sqrt{36}*\sqrt{473}=6\sqrt{473}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{473}}{2*18}=\frac{-6-6\sqrt{473}}{36}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{473}}{2*18}=\frac{-6+6\sqrt{473}}{36}
$