7j(j-3)=4(3-j)

Solution for 7j(j-3)=4(3-j) equation:

7j(j-3)=4(3-j)

We move all terms to the left:

7j(j-3)-(4(3-j))=0

We add all the numbers together, and all the variables

7j(j-3)-(4(-1j+3))=0

We multiply parentheses

7j^2-21j-(4(-1j+3))=0


We calculate terms in parentheses: -(4(-1j+3)), so:

4(-1j+3)

We multiply parentheses

-4j+12

Back to the equation:

-(-4j+12)


We get rid of parentheses

7j^2-21j+4j-12=0

We add all the numbers together, and all the variables

7j^2-17j-12=0

a = 7; b = -17; c = -12;
Δ = b2-4ac
Δ = -172-4·7·(-12)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-25}{2*7}=\frac{-8}{14}
=-4/7
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+25}{2*7}=\frac{42}{14}
=3
$