q=1+3/4q

Solution for q=1+3/4q equation:

q=1+3/4q

We move all terms to the left:

q-(1+3/4q)=0


Domain of the equation: 4q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

q-(3/4q+1)=0

We get rid of parentheses

q-3/4q-1=0

We multiply all the terms by the denominator


q*4q-1*4q-3=0

Wy multiply elements

4q^2-4q-3=0

a = 4; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·4·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*4}=\frac{-4}{8}
=-1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*4}=\frac{12}{8}
=1+1/2
$