(c-6)(c+2)=-7

Solution for (c-6)(c+2)=-7 equation:

(c-6)(c+2)=-7

We move all terms to the left:

(c-6)(c+2)-(-7)=0

We add all the numbers together, and all the variables

(c-6)(c+2)+7=0

We multiply parentheses ..

(+c^2+2c-6c-12)+7=0

We get rid of parentheses

c^2+2c-6c-12+7=0

We add all the numbers together, and all the variables

c^2-4c-5=0

a = 1; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·1·(-5)
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{-(-4)-6}{2*1}=\frac{-2}{2}
=-1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*1}=\frac{10}{2}
=5
$