(T-3)(t-3)-(t-2)(t-2)=5

Solution for (T-3)(t-3)-(t-2)(t-2)=5 equation:

(-3)(T-3)-(T-2)(T-2)=5

We move all terms to the left:

(-3)(T-3)-(T-2)(T-2)-(5)=0

We multiply parentheses ..

(-3T+9)-(T-2)(T-2)-5=0

We get rid of parentheses

-3T-(T-2)(T-2)+9-5=0

We multiply parentheses ..

-(+T^2-2T-2T+4)-3T+9-5=0

We add all the numbers together, and all the variables

-(+T^2-2T-2T+4)-3T+4=0

We get rid of parentheses

-T^2+2T+2T-3T-4+4=0

We add all the numbers together, and all the variables

-1T^2+T=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
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{-(1)-1}{2*-1}=\frac{-2}{-2}
=1
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-1}=\frac{0}{-2}
=0
$