-16j^2+64j=48

Solution for -16j^2+64j=48 equation:

-16j^2+64j=48

We move all terms to the left:

-16j^2+64j-(48)=0

a = -16; b = 64; c = -48;
Δ = b2-4ac
Δ = 642-4·(-16)·(-48)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32}{2*-16}=\frac{-96}{-32}
=+3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32}{2*-16}=\frac{-32}{-32}
=1
$