Y=3x^2+24x+42

Solution for Y=3x^2+24x+42 equation:

=3Y^2+24Y+42

We move all terms to the left:

-(3Y^2+24Y+42)=0

We get rid of parentheses

-3Y^2-24Y-42=0

a = -3; b = -24; c = -42;
Δ = b2-4ac
Δ = -242-4·(-3)·(-42)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{2}}{2*-3}=\frac{24-6\sqrt{2}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{2}}{2*-3}=\frac{24+6\sqrt{2}}{-6}
$