(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 get rid of parentheses

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

We multiply parentheses ..

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

We calculate fractions


((+3p^2+3*20)/()+()/()=0


We calculate terms in parentheses: +((+3p^2+3*20)/()+()/(), so:

(+3p^2+3*20)/()+()/(

We add all the numbers together, and all the variables

(+3p^2+3*20)/()+1

We multiply all the terms by the denominator


(+3p^2+3*20)+1*()

We add all the numbers together, and all the variables

(+3p^2+3*20)

We get rid of parentheses

3p^2+3*20

We add all the numbers together, and all the variables

3p^2+60

Back to the equation:

+(3p^2+60)


We get rid of parentheses

3p^2+60=0

a = 3; b = 0; c = +60;
Δ = b2-4ac
Δ = 02-4·3·60
Δ = -720
Delta is less than zero, so there is no solution for the equation