r(3+4r)=20

Solution for r(3+4r)=20 equation:

r(3+4r)=20

We move all terms to the left:

r(3+4r)-(20)=0

We add all the numbers together, and all the variables

r(4r+3)-20=0

We multiply parentheses

4r^2+3r-20=0

a = 4; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·4·(-20)
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{329}}{2*4}=\frac{-3-\sqrt{329}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{329}}{2*4}=\frac{-3+\sqrt{329}}{8}
$