(c+2)(c-5)=-6

Solution for (c+2)(c-5)=-6 equation:

(c+2)(c-5)=-6

We move all terms to the left:

(c+2)(c-5)-(-6)=0

We add all the numbers together, and all the variables

(c+2)(c-5)+6=0

We multiply parentheses ..

(+c^2-5c+2c-10)+6=0

We get rid of parentheses

c^2-5c+2c-10+6=0

We add all the numbers together, and all the variables

c^2-3c-4=0

a = 1; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*1}=\frac{-2}{2}
=-1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*1}=\frac{8}{2}
=4
$