12t^2-18t+6=0

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

12t^2-18t+6=0

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

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