(3/5)r=42

Solution for (3/5)r=42 equation:

(3/5)r=42

We move all terms to the left:

(3/5)r-(42)=0


Domain of the equation: 5)r!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

(+3/5)r-42=0

We multiply parentheses

3r^2-42=0

a = 3; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·3·(-42)
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*3}=\frac{0-6\sqrt{14}}{6}
=-\frac{6\sqrt{14}}{6}
=-\sqrt{14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*3}=\frac{0+6\sqrt{14}}{6}
=\frac{6\sqrt{14}}{6}
=\sqrt{14}
$