Y=3x^2+22x+24

Solution for Y=3x^2+22x+24 equation:

=3Y^2+22Y+24

We move all terms to the left:

-(3Y^2+22Y+24)=0

We get rid of parentheses

-3Y^2-22Y-24=0

a = -3; b = -22; c = -24;
Δ = b2-4ac
Δ = -222-4·(-3)·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*-3}=\frac{8}{-6}
=-1+1/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*-3}=\frac{36}{-6}
=-6
$