2q+1=(100+q2+q)/q

Solution for 2q+1=(100+q2+q)/q equation:

2q+1=(100+q2+q)/q

We move all terms to the left:

2q+1-((100+q2+q)/q)=0


Domain of the equation: q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

-((+q^2+q+100)/q)+2q+1=0

We multiply all the terms by the denominator


-((+q^2+q+100)+2q*q)+1*q)=0


We calculate terms in parentheses: -((+q^2+q+100)+2q*q), so:

(+q^2+q+100)+2q*q

Wy multiply elements

(+q^2+q+100)+2q^2

We get rid of parentheses

q^2+2q^2+q+100

We add all the numbers together, and all the variables

3q^2+q+100

Back to the equation:

-(3q^2+q+100)


Wy multiply elements

-(3q^2+q+100)+q=0

We get rid of parentheses

-3q^2-q+q-100=0

We add all the numbers together, and all the variables

-3q^2-100=0

a = -3; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·(-3)·(-100)
Δ = -1200
Delta is less than zero, so there is no solution for the equation