3r^2-19r+28=9

Solution for 3r^2-19r+28=9 equation:

3r^2-19r+28=9

We move all terms to the left:

3r^2-19r+28-(9)=0

We add all the numbers together, and all the variables

3r^2-19r+19=0

a = 3; b = -19; c = +19;
Δ = b2-4ac
Δ = -192-4·3·19
Δ = 133
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{-(-19)-\sqrt{133}}{2*3}=\frac{19-\sqrt{133}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{133}}{2*3}=\frac{19+\sqrt{133}}{6}
$