(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^2-15c-6=0

a = 1; b = -15; c = -6;
Δ = b2-4ac
Δ = -152-4·1·(-6)
Δ = 249
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{-(-15)-\sqrt{249}}{2*1}=\frac{15-\sqrt{249}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{249}}{2*1}=\frac{15+\sqrt{249}}{2}
$