(2t-6)(3t-2)=0

Solution for (2t-6)(3t-2)=0 equation:

(2t-6)(3t-2)=0

We multiply parentheses ..

(+6t^2-4t-18t+12)=0

We get rid of parentheses

6t^2-4t-18t+12=0

We add all the numbers together, and all the variables

6t^2-22t+12=0

a = 6; b = -22; c = +12;
Δ = b2-4ac
Δ = -222-4·6·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*6}=\frac{8}{12}
=2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*6}=\frac{36}{12}
=3
$