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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

t^2-5t+6=0

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

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