Y=-3x^2+18x-23

Solution for Y=-3x^2+18x-23 equation:

=-3Y^2+18Y-23

We move all terms to the left:

-(-3Y^2+18Y-23)=0

We get rid of parentheses

3Y^2-18Y+23=0

a = 3; b = -18; c = +23;
Δ = b2-4ac
Δ = -182-4·3·23
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{3}}{2*3}=\frac{18-4\sqrt{3}}{6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{3}}{2*3}=\frac{18+4\sqrt{3}}{6}
$