(2t+1)(t-4)=t^-4t-6

Solution for (2t+1)(t-4)=t^-4t-6 equation:

(2t+1)(t-4)=t^-4t-6

We move all terms to the left:

(2t+1)(t-4)-(t^-4t-6)=0

We get rid of parentheses

(2t+1)(t-4)-t^+4t+6=0

We multiply parentheses ..

(+2t^2-8t+t-4)-t^+4t+6=0

We add all the numbers together, and all the variables

(+2t^2-8t+t-4)+3t+6=0

We get rid of parentheses

2t^2-8t+t+3t-4+6=0

We add all the numbers together, and all the variables

2t^2-4t+2=0

a = 2; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·2·2
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{4}{4}=1$