(c+2)(c-1)=28

Solution for (c+2)(c-1)=28 equation:

(c+2)(c-1)=28

We move all terms to the left:

(c+2)(c-1)-(28)=0

We multiply parentheses ..

(+c^2-1c+2c-2)-28=0

We get rid of parentheses

c^2-1c+2c-2-28=0

We add all the numbers together, and all the variables

c^2+c-30=0

a = 1; b = 1; c = -30;
Δ = b2-4ac
Δ = 12-4·1·(-30)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*1}=\frac{-12}{2}
=-6
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*1}=\frac{10}{2}
=5
$