(8j+5)(j+2)=0

Solution for (8j+5)(j+2)=0 equation:

(8j+5)(j+2)=0

We multiply parentheses ..

(+8j^2+16j+5j+10)=0

We get rid of parentheses

8j^2+16j+5j+10=0

We add all the numbers together, and all the variables

8j^2+21j+10=0

a = 8; b = 21; c = +10;
Δ = b2-4ac
Δ = 212-4·8·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*8}=\frac{-32}{16}
=-2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*8}=\frac{-10}{16}
=-5/8
$