(t-2)(2t-9)=0

Solution for (t-2)(2t-9)=0 equation:

(t-2)(2t-9)=0

We multiply parentheses ..

(+2t^2-9t-4t+18)=0

We get rid of parentheses

2t^2-9t-4t+18=0

We add all the numbers together, and all the variables

2t^2-13t+18=0

a = 2; b = -13; c = +18;
Δ = b2-4ac
Δ = -132-4·2·18
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{-(-13)-5}{2*2}=\frac{8}{4}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*2}=\frac{18}{4}
=4+1/2
$