3y^2-19y+28=0

Solution for 3y^2-19y+28=0 equation:

3y^2-19y+28=0

a = 3; b = -19; c = +28;
Δ = b2-4ac
Δ = -192-4·3·28
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*3}=\frac{14}{6}
=2+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*3}=\frac{24}{6}
=4
$