(2j-3)(j-8)=0

Solution for (2j-3)(j-8)=0 equation:

(2j-3)(j-8)=0

We multiply parentheses ..

(+2j^2-16j-3j+24)=0

We get rid of parentheses

2j^2-16j-3j+24=0

We add all the numbers together, and all the variables

2j^2-19j+24=0

a = 2; b = -19; c = +24;
Δ = b2-4ac
Δ = -192-4·2·24
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*2}=\frac{6}{4}
=1+1/2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*2}=\frac{32}{4}
=8
$