(4(r))-18=20+(1/5(r))

Solution for (4(r))-18=20+(1/5(r)) equation:

(4(r))-18=20+(1/5(r))

We move all terms to the left:

(4(r))-18-(20+(1/5(r)))=0


Domain of the equation: 5r))!=0

r!=0/1

r!=0

r∈R


determiningTheFunctionDomain
4r-(20+(1/5r))-18=0

We add all the numbers together, and all the variables

4r-(20+(+1/5r))-18=0

We multiply all the terms by the denominator


4r*5r))-(20+(-18*5r))+1=0

Wy multiply elements

20r^2-90r=0

a = 20; b = -90; c = 0;
Δ = b2-4ac
Δ = -902-4·20·0
Δ = 8100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8100}=90$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-90}{2*20}=\frac{0}{40}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+90}{2*20}=\frac{180}{40}
=4+1/2
$