(1/5)(k-3)=(3/4)

Solution for (1/5)(k-3)=(3/4) equation:

(1/5)(k-3)=(3/4)

We move all terms to the left:

(1/5)(k-3)-((3/4))=0


Domain of the equation: 5)(k-3)!=0

k∈R


We add all the numbers together, and all the variables

(+1/5)(k-3)-((+3/4))=0

We multiply parentheses ..

(+k^2+1/5*-3)-((+3/4))=0

We calculate fractions


k^2/()+()/()=0

We add all the numbers together, and all the variables

k^2/()+1=0

We multiply all the terms by the denominator


k^2+1*()=0

We add all the numbers together, and all the variables

k^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

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