5(r-3)=10((r/2)-3)

Solution for 5(r-3)=10((r/2)-3) equation:

5(r-3)=10((r/2)-3)

We move all terms to the left:

5(r-3)-(10((r/2)-3))=0

We add all the numbers together, and all the variables

5(r-3)-(10((+r/2)-3))=0

We multiply parentheses

5r-(10((+r/2)-3))-15=0

We multiply all the terms by the denominator


5r*2)-3))-(10((+r-15*2)-3))=0

We add all the numbers together, and all the variables

5r*2)-3))-(10((r-30)-3))=0

We add all the numbers together, and all the variables

5r*2)-3))-(10((r=0

Wy multiply elements

10r^2=0

a = 10; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·10·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$r=\frac{-b}{2a}=\frac{0}{20}=0$