(3/5)(1+p)=(21/20)

Solution for (3/5)(1+p)=(21/20) equation:

(3/5)(1+p)=(21/20)

We move all terms to the left:

(3/5)(1+p)-((21/20))=0


Domain of the equation: 5)(1+p)!=0

p∈R


We add all the numbers together, and all the variables

(+3/5)(p+1)-((+21/20))=0

We multiply parentheses ..

(+3p^2+3/5*1)-((+21/20))=0

We calculate fractions


3p^2/()+()/()=0

We add all the numbers together, and all the variables

3p^2/()+1=0

We multiply all the terms by the denominator


3p^2+1*()=0

We add all the numbers together, and all the variables

3p^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:
$p=\frac{-b}{2a}=\frac{0}{6}=0$