(-2/5)*j+2/5=0

Solution for (-2/5)*j+2/5=0 equation:

(-2/5)*j+2/5=0


Domain of the equation: 5)*j!=0

j!=0/1

j!=0

j∈R


We multiply parentheses

-2j^2+2/5=0

We multiply all the terms by the denominator


-2j^2*5+2=0

Wy multiply elements

-10j^2+2=0

a = -10; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-10)·2
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-10}=\frac{0-4\sqrt{5}}{-20}
=-\frac{4\sqrt{5}}{-20}
=-\frac{\sqrt{5}}{-5}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-10}=\frac{0+4\sqrt{5}}{-20}
=\frac{4\sqrt{5}}{-20}
=\frac{\sqrt{5}}{-5}
$