64=j^2=0

Solution for 64=j^2=0 equation:

64=j^2=0

We move all terms to the left:

64-(j^2)=0

We add all the numbers together, and all the variables

-1j^2+64=0

a = -1; b = 0; c = +64;
Δ = b2-4ac
Δ = 02-4·(-1)·64
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-1}=\frac{-16}{-2}
=+8
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-1}=\frac{16}{-2}
=-8
$