(c+1)(c-3)=21

Solution for (c+1)(c-3)=21 equation:

(c+1)(c-3)=21

We move all terms to the left:

(c+1)(c-3)-(21)=0

We multiply parentheses ..

(+c^2-3c+c-3)-21=0

We get rid of parentheses

c^2-3c+c-3-21=0

We add all the numbers together, and all the variables

c^2-2c-24=0

a = 1; b = -2; c = -24;
Δ = b2-4ac
Δ = -22-4·1·(-24)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*1}=\frac{-8}{2}
=-4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*1}=\frac{12}{2}
=6
$