Y=3x^2+42x+138

Solution for Y=3x^2+42x+138 equation:

=3Y^2+42Y+138

We move all terms to the left:

-(3Y^2+42Y+138)=0

We get rid of parentheses

-3Y^2-42Y-138=0

a = -3; b = -42; c = -138;
Δ = b2-4ac
Δ = -422-4·(-3)·(-138)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{3}}{2*-3}=\frac{42-6\sqrt{3}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{3}}{2*-3}=\frac{42+6\sqrt{3}}{-6}
$