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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

c^2-1c-20=0

a = 1; b = -1; c = -20;
Δ = b2-4ac
Δ = -12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*1}=\frac{-8}{2}
=-4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*1}=\frac{10}{2}
=5
$