j^2+3j-12=6

Solution for j^2+3j-12=6 equation:

j^2+3j-12=6

We move all terms to the left:

j^2+3j-12-(6)=0

We add all the numbers together, and all the variables

j^2+3j-18=0

a = 1; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*1}=\frac{-12}{2}
=-6
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*1}=\frac{6}{2}
=3
$