3r^2=11r+20

Solution for 3r^2=11r+20 equation:

3r^2=11r+20

We move all terms to the left:

3r^2-(11r+20)=0

We get rid of parentheses

3r^2-11r-20=0

a = 3; b = -11; c = -20;
Δ = b2-4ac
Δ = -112-4·3·(-20)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*3}=\frac{-8}{6}
=-1+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*3}=\frac{30}{6}
=5
$