3r^2+20=17r

Solution for 3r^2+20=17r equation:

3r^2+20=17r

We move all terms to the left:

3r^2+20-(17r)=0

a = 3; b = -17; c = +20;
Δ = b2-4ac
Δ = -172-4·3·20
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-7}{2*3}=\frac{10}{6}
=1+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+7}{2*3}=\frac{24}{6}
=4
$