-j2=-62

Solution for -j2=-62 equation:

-j2=-62

We move all terms to the left:

-j2-(-62)=0

We add all the numbers together, and all the variables

-1j^2+62=0

a = -1; b = 0; c = +62;
Δ = b2-4ac
Δ = 02-4·(-1)·62
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{62}}{2*-1}=\frac{0-2\sqrt{62}}{-2}
=-\frac{2\sqrt{62}}{-2}
=-\frac{\sqrt{62}}{-1}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{62}}{2*-1}=\frac{0+2\sqrt{62}}{-2}
=\frac{2\sqrt{62}}{-2}
=\frac{\sqrt{62}}{-1}
$