(2t-7)(t-3)=0

Solution for (2t-7)(t-3)=0 equation:

(2t-7)(t-3)=0

We multiply parentheses ..

(+2t^2-6t-7t+21)=0

We get rid of parentheses

2t^2-6t-7t+21=0

We add all the numbers together, and all the variables

2t^2-13t+21=0

a = 2; b = -13; c = +21;
Δ = b2-4ac
Δ = -132-4·2·21
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{-(-13)-1}{2*2}=\frac{12}{4}
=3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*2}=\frac{14}{4}
=3+1/2
$