(j+1)(j-2)=0

Solution for (j+1)(j-2)=0 equation:

(j+1)(j-2)=0

We multiply parentheses ..

(+j^2-2j+j-2)=0

We get rid of parentheses

j^2-2j+j-2=0

We add all the numbers together, and all the variables

j^2-1j-2=0

a = 1; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*1}=\frac{-2}{2}
=-1
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*1}=\frac{4}{2}
=2
$