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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -((t-2)+2(t-3)), so:

(t-2)+2(t-3)

We multiply parentheses

(t-2)+2t-6

We get rid of parentheses

t+2t-2-6

We add all the numbers together, and all the variables

3t-8

Back to the equation:

-(3t-8)


We get rid of parentheses

t^2-3t-3t+8=0

We add all the numbers together, and all the variables

t^2-6t+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2}{2*1}=\frac{4}{2}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*1}=\frac{8}{2}
=4
$