6y^2-18y+12=0

Solution for 6y^2-18y+12=0 equation:

6y^2-18y+12=0

a = 6; b = -18; c = +12;
Δ = b2-4ac
Δ = -182-4·6·12
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*6}=\frac{12}{12}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*6}=\frac{24}{12}
=2
$