16=j2+12

Solution for 16=j2+12 equation:

16=j2+12

We move all terms to the left:

16-(j2+12)=0

We add all the numbers together, and all the variables

-(+j^2+12)+16=0

We get rid of parentheses

-j^2-12+16=0

We add all the numbers together, and all the variables

-1j^2+4=0

a = -1; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-1)·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-1}=\frac{-4}{-2}
=+2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-1}=\frac{4}{-2}
=-2
$