Y=3x^2-19x+28

Solution for Y=3x^2-19x+28 equation:

=3Y^2-19Y+28

We move all terms to the left:

-(3Y^2-19Y+28)=0

We get rid of parentheses

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