Y=12x^2-19x+5

Solution for Y=12x^2-19x+5 equation:

=12Y^2-19Y+5

We move all terms to the left:

-(12Y^2-19Y+5)=0

We get rid of parentheses

-12Y^2+19Y-5=0

a = -12; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·(-12)·(-5)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*-12}=\frac{-30}{-24}
=1+1/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*-12}=\frac{-8}{-24}
=1/3
$