(t-3)(t-9)=0

Solution for (t-3)(t-9)=0 equation:

(t-3)(t-9)=0

We multiply parentheses ..

(+t^2-9t-3t+27)=0

We get rid of parentheses

t^2-9t-3t+27=0

We add all the numbers together, and all the variables

t^2-12t+27=0

a = 1; b = -12; c = +27;
Δ = b2-4ac
Δ = -122-4·1·27
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{-(-12)-6}{2*1}=\frac{6}{2}
=3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6}{2*1}=\frac{18}{2}
=9
$