12j^2=32j-12

Solution for 12j^2=32j-12 equation:

12j^2=32j-12

We move all terms to the left:

12j^2-(32j-12)=0

We get rid of parentheses

12j^2-32j+12=0

a = 12; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·12·12
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{7}}{2*12}=\frac{32-8\sqrt{7}}{24}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{7}}{2*12}=\frac{32+8\sqrt{7}}{24}
$