r+3=(10-r)/3

Solution for r+3=(10-r)/3 equation:

r+3=(10-r)/3

We move all terms to the left:

r+3-((10-r)/3)=0

We add all the numbers together, and all the variables

r-((-1r+10)/3)+3=0

We multiply all the terms by the denominator


r*3)-((-1r+10)+3*3)=0

We add all the numbers together, and all the variables

r*3)-((-1r+10)=0

Wy multiply elements

3r^2=0

a = 3; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·3·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$r=\frac{-b}{2a}=\frac{0}{6}=0$