Y=3x^2-24x+22

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

=3Y^2-24Y+22

We move all terms to the left:

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

We get rid of parentheses

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

a = -3; b = 24; c = -22;
Δ = b2-4ac
Δ = 242-4·(-3)·(-22)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{78}}{2*-3}=\frac{-24-2\sqrt{78}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{78}}{2*-3}=\frac{-24+2\sqrt{78}}{-6}
$