14=j2+12

Solution for 14=j2+12 equation:

14=j2+12

We move all terms to the left:

14-(j2+12)=0

We add all the numbers together, and all the variables

-(+j^2+12)+14=0

We get rid of parentheses

-j^2-12+14=0

We add all the numbers together, and all the variables

-1j^2+2=0

a = -1; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-1)·2
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-1}=\frac{0-2\sqrt{2}}{-2}
=-\frac{2\sqrt{2}}{-2}
=-\frac{\sqrt{2}}{-1}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-1}=\frac{0+2\sqrt{2}}{-2}
=\frac{2\sqrt{2}}{-2}
=\frac{\sqrt{2}}{-1}
$