7/3q+16=5q

Solution for 7/3q+16=5q equation:

7/3q+16=5q

We move all terms to the left:

7/3q+16-(5q)=0


Domain of the equation: 3q!=0

q!=0/3

q!=0

q∈R


We add all the numbers together, and all the variables

-5q+7/3q+16=0

We multiply all the terms by the denominator


-5q*3q+16*3q+7=0

Wy multiply elements

-15q^2+48q+7=0

a = -15; b = 48; c = +7;
Δ = b2-4ac
Δ = 482-4·(-15)·7
Δ = 2724
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:

$\sqrt{\Delta}=\sqrt{2724}=\sqrt{4*681}=\sqrt{4}*\sqrt{681}=2\sqrt{681}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-2\sqrt{681}}{2*-15}=\frac{-48-2\sqrt{681}}{-30}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+2\sqrt{681}}{2*-15}=\frac{-48+2\sqrt{681}}{-30}
$