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

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

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

We multiply parentheses ..

(+6t^2+4t-9t-6)=0

We get rid of parentheses

6t^2+4t-9t-6=0

We add all the numbers together, and all the variables

6t^2-5t-6=0

a = 6; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·6·(-6)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*6}=\frac{-8}{12}
=-2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*6}=\frac{18}{12}
=1+1/2
$