4/5q=32q=40

Solution for 4/5q=32q=40 equation:

4/5q=32q=40

We move all terms to the left:

4/5q-(32q)=0


Domain of the equation: 5q!=0

q!=0/5

q!=0

q∈R


We add all the numbers together, and all the variables

-32q+4/5q=0

We multiply all the terms by the denominator


-32q*5q+4=0

Wy multiply elements

-160q^2+4=0

a = -160; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-160)·4
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*-160}=\frac{0-16\sqrt{10}}{-320}
=-\frac{16\sqrt{10}}{-320}
=-\frac{\sqrt{10}}{-20}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*-160}=\frac{0+16\sqrt{10}}{-320}
=\frac{16\sqrt{10}}{-320}
=\frac{\sqrt{10}}{-20}
$