(2t+2)(t-6)=(t-2)(t-2)

Solution for (2t+2)(t-6)=(t-2)(t-2) equation:

(2t+2)(t-6)=(t-2)(t-2)

We move all terms to the left:

(2t+2)(t-6)-((t-2)(t-2))=0

We multiply parentheses ..

(+2t^2-12t+2t-12)-((t-2)(t-2))=0


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

(t-2)(t-2)

We multiply parentheses ..

(+t^2-2t-2t+4)

We get rid of parentheses

t^2-2t-2t+4

We add all the numbers together, and all the variables

t^2-4t+4

Back to the equation:

-(t^2-4t+4)


We get rid of parentheses

2t^2-t^2-12t+2t+4t-12-4=0

We add all the numbers together, and all the variables

t^2-6t-16=0

a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2}
=-2$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2}
=8$