21r(63r+3)=3/7

Solution for 21r(63r+3)=3/7 equation:

21r(63r+3)=3/7

We move all terms to the left:

21r(63r+3)-(3/7)=0

We add all the numbers together, and all the variables

21r(63r+3)-(+3/7)=0

We multiply parentheses

1323r^2+63r-(+3/7)=0

We get rid of parentheses

1323r^2+63r-3/7=0

We multiply all the terms by the denominator


1323r^2*7+63r*7-3=0

Wy multiply elements

9261r^2+441r-3=0

a = 9261; b = 441; c = -3;
Δ = b2-4ac
Δ = 4412-4·9261·(-3)
Δ = 305613
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:

$\sqrt{\Delta}=\sqrt{305613}=\sqrt{3969*77}=\sqrt{3969}*\sqrt{77}=63\sqrt{77}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(441)-63\sqrt{77}}{2*9261}=\frac{-441-63\sqrt{77}}{18522}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(441)+63\sqrt{77}}{2*9261}=\frac{-441+63\sqrt{77}}{18522}
$