Y=3x^2-24x+43

Solution for Y=3x^2-24x+43 equation:

=3Y^2-24Y+43

We move all terms to the left:

-(3Y^2-24Y+43)=0

We get rid of parentheses

-3Y^2+24Y-43=0

a = -3; b = 24; c = -43;
Δ = b2-4ac
Δ = 242-4·(-3)·(-43)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{15}}{2*-3}=\frac{-24-2\sqrt{15}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{15}}{2*-3}=\frac{-24+2\sqrt{15}}{-6}
$