64j^2-128j+64=39

Solution for 64j^2-128j+64=39 equation:

64j^2-128j+64=39

We move all terms to the left:

64j^2-128j+64-(39)=0

We add all the numbers together, and all the variables

64j^2-128j+25=0

a = 64; b = -128; c = +25;
Δ = b2-4ac
Δ = -1282-4·64·25
Δ = 9984
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:

$\sqrt{\Delta}=\sqrt{9984}=\sqrt{256*39}=\sqrt{256}*\sqrt{39}=16\sqrt{39}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-16\sqrt{39}}{2*64}=\frac{128-16\sqrt{39}}{128}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+16\sqrt{39}}{2*64}=\frac{128+16\sqrt{39}}{128}
$