(3t-1)(2t+1)=0

Solution for (3t-1)(2t+1)=0 equation:

(3t-1)(2t+1)=0

We multiply parentheses ..

(+6t^2+3t-2t-1)=0

We get rid of parentheses

6t^2+3t-2t-1=0

We add all the numbers together, and all the variables

6t^2+t-1=0

a = 6; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·6·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*6}=\frac{-6}{12}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*6}=\frac{4}{12}
=1/3
$