(c+2)(c+3)=14c

Solution for (c+2)(c+3)=14c equation:

(c+2)(c+3)=14c

We move all terms to the left:

(c+2)(c+3)-(14c)=0

We add all the numbers together, and all the variables

-14c+(c+2)(c+3)=0

We multiply parentheses ..

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

We get rid of parentheses

c^2+3c+2c-14c+6=0

We add all the numbers together, and all the variables

c^2-9c+6=0

a = 1; b = -9; c = +6;
Δ = b2-4ac
Δ = -92-4·1·6
Δ = 57
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{57}}{2*1}=\frac{9-\sqrt{57}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{57}}{2*1}=\frac{9+\sqrt{57}}{2}
$