27/4r=5r

Solution for 27/4r=5r equation:

27/4r=5r

We move all terms to the left:

27/4r-(5r)=0


Domain of the equation: 4r!=0

r!=0/4

r!=0

r∈R


We add all the numbers together, and all the variables

-5r+27/4r=0

We multiply all the terms by the denominator


-5r*4r+27=0

Wy multiply elements

-20r^2+27=0

a = -20; b = 0; c = +27;
Δ = b2-4ac
Δ = 02-4·(-20)·27
Δ = 2160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{15}}{2*-20}=\frac{0-12\sqrt{15}}{-40}
=-\frac{12\sqrt{15}}{-40}
=-\frac{3\sqrt{15}}{-10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{15}}{2*-20}=\frac{0+12\sqrt{15}}{-40}
=\frac{12\sqrt{15}}{-40}
=\frac{3\sqrt{15}}{-10}
$