7-(1/4)(j-8)=6

Solution for 7-(1/4)(j-8)=6 equation:

7-(1/4)(j-8)=6

We move all terms to the left:

7-(1/4)(j-8)-(6)=0


Domain of the equation: 4)(j-8)!=0

j∈R


We add all the numbers together, and all the variables

-(+1/4)(j-8)+7-6=0

We add all the numbers together, and all the variables

-(+1/4)(j-8)+1=0

We multiply parentheses ..

-(+j^2+1/4*-8)+1=0

We multiply all the terms by the denominator


-(+j^2+1+1*4*-8)=0

We get rid of parentheses

-j^2-1+8-1*4*=0

We add all the numbers together, and all the variables

-1j^2=0

a = -1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-1)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$j=\frac{-b}{2a}=\frac{0}{-2}=0$