1/3q+q=4

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

1/3q+q=4

We move all terms to the left:

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


Domain of the equation: 3q!=0

q!=0/3

q!=0

q∈R


We add all the numbers together, and all the variables

q+1/3q-4=0

We multiply all the terms by the denominator


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

Wy multiply elements

3q^2-12q+1=0

a = 3; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·3·1
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{33}}{2*3}=\frac{12-2\sqrt{33}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{33}}{2*3}=\frac{12+2\sqrt{33}}{6}
$