4j(j-3)=84

Solution for 4j(j-3)=84 equation:

4j(j-3)=84

We move all terms to the left:

4j(j-3)-(84)=0

We multiply parentheses

4j^2-12j-84=0

a = 4; b = -12; c = -84;
Δ = b2-4ac
Δ = -122-4·4·(-84)
Δ = 1488
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:

$\sqrt{\Delta}=\sqrt{1488}=\sqrt{16*93}=\sqrt{16}*\sqrt{93}=4\sqrt{93}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{93}}{2*4}=\frac{12-4\sqrt{93}}{8}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{93}}{2*4}=\frac{12+4\sqrt{93}}{8}
$