(k+2)(k-1)=3

Solution for (k+2)(k-1)=3 equation:

(k+2)(k-1)=3

We move all terms to the left:

(k+2)(k-1)-(3)=0

We multiply parentheses ..

(+k^2-1k+2k-2)-3=0

We get rid of parentheses

k^2-1k+2k-2-3=0

We add all the numbers together, and all the variables

k^2+k-5=0

a = 1; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·1·(-5)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{21}}{2*1}=\frac{-1-\sqrt{21}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21}}{2*1}=\frac{-1+\sqrt{21}}{2}
$