/k^2-4=29

Solution for /k^2-4=29 equation:

/k^2-4=29

We move all terms to the left:

/k^2-4-(29)=0


Domain of the equation: k^2!=0

k^2!=0/

k^2!=√0

k!=0

k∈R


We add all the numbers together, and all the variables

/k^2-33=0

We multiply all the terms by the denominator


-33*k^2+=0

We add all the numbers together, and all the variables

-33k^2=0

a = -33; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-33)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$k=\frac{-b}{2a}=\frac{0}{-66}=0$