-24y^2+48y-18=0

Solution for -24y^2+48y-18=0 equation:

-24y^2+48y-18=0

a = -24; b = 48; c = -18;
Δ = b2-4ac
Δ = 482-4·(-24)·(-18)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-24}{2*-24}=\frac{-72}{-48}
=1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+24}{2*-24}=\frac{-24}{-48}
=1/2
$