212q+1=(1/2q+1)=-3(2q-1)+4(2q+1)

Solution for 212q+1=(1/2q+1)=-3(2q-1)+4(2q+1) equation:

212q+1=(1/2q+1)=-3(2q-1)+4(2q+1)

We move all terms to the left:

212q+1-((1/2q+1))=0


Domain of the equation: 2q+1))!=0

q∈R


We multiply all the terms by the denominator


212q*2q+1*2q+1))-((1+1))=0

We add all the numbers together, and all the variables

212q*2q+1*2q+1))-(2)=0

We add all the numbers together, and all the variables

212q*2q+1*2q=0

Wy multiply elements

424q^2+2q=0

a = 424; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·424·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*424}=\frac{-4}{848}
=-1/212
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*424}=\frac{0}{848}
=0
$