Y=3x^2-42x+5

Solution for Y=3x^2-42x+5 equation:

=3Y^2-42Y+5

We move all terms to the left:

-(3Y^2-42Y+5)=0

We get rid of parentheses

-3Y^2+42Y-5=0

a = -3; b = 42; c = -5;
Δ = b2-4ac
Δ = 422-4·(-3)·(-5)
Δ = 1704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1704}=\sqrt{4*426}=\sqrt{4}*\sqrt{426}=2\sqrt{426}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{426}}{2*-3}=\frac{-42-2\sqrt{426}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{426}}{2*-3}=\frac{-42+2\sqrt{426}}{-6}
$