10j-6(2j-1)=-2j(j-3)

Solution for 10j-6(2j-1)=-2j(j-3) equation:

10j-6(2j-1)=-2j(j-3)

We move all terms to the left:

10j-6(2j-1)-(-2j(j-3))=0

We multiply parentheses

10j-12j-(-2j(j-3))+6=0


We calculate terms in parentheses: -(-2j(j-3)), so:

-2j(j-3)

We multiply parentheses

-2j^2+6j

Back to the equation:

-(-2j^2+6j)


We add all the numbers together, and all the variables

-(-2j^2+6j)-2j+6=0

We get rid of parentheses

2j^2-6j-2j+6=0

We add all the numbers together, and all the variables

2j^2-8j+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$