7j(2j+5)=0

Solution for 7j(2j+5)=0 equation:

7j(2j+5)=0

We multiply parentheses

14j^2+35j=0

a = 14; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·14·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*14}=\frac{-70}{28}
=-2+1/2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*14}=\frac{0}{28}
=0
$