(2t-1)(t-3)=t-3

Solution for (2t-1)(t-3)=t-3 equation:

(2t-1)(t-3)=t-3

We move all terms to the left:

(2t-1)(t-3)-(t-3)=0

We get rid of parentheses

(2t-1)(t-3)-t+3=0

We multiply parentheses ..

(+2t^2-6t-1t+3)-t+3=0

We add all the numbers together, and all the variables

(+2t^2-6t-1t+3)-1t+3=0

We get rid of parentheses

2t^2-6t-1t-1t+3+3=0

We add all the numbers together, and all the variables

2t^2-8t+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$