-(2/3)(5p-16)=10

Solution for -(2/3)(5p-16)=10 equation:

-(2/3)(5p-16)=10

We move all terms to the left:

-(2/3)(5p-16)-(10)=0


Domain of the equation: 3)(5p-16)!=0

p∈R


We add all the numbers together, and all the variables

-(+2/3)(5p-16)-10=0

We multiply parentheses ..

-(+10p^2+2/3*-16)-10=0

We multiply all the terms by the denominator


-(+10p^2+2-10*3*-16)=0

We get rid of parentheses

-10p^2-2+16+10*3*=0

We add all the numbers together, and all the variables

-10p^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:
$p=\frac{-b}{2a}=\frac{0}{-20}=0$