(1/4)t^2+t+1=0

Solution for (1/4)t^2+t+1=0 equation:

(1/4)t^2+t+1=0


Domain of the equation: 4)t^2!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+1/4)t^2+t+1=0

We add all the numbers together, and all the variables

t+(+1/4)t^2+1=0

We multiply all the terms by the denominator


t*4)t^2+(+1*4)t^2+1=0

We add all the numbers together, and all the variables

t*4)t^2+4t^2+1=0

Wy multiply elements

4t^2=0

a = 4; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·4·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{0}{8}=0$