(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 get rid of parentheses

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

We multiply parentheses ..

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

We calculate fractions


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


We calculate terms in parentheses: +((+k^2+1*4)/()+()/(), so:

(+k^2+1*4)/()+()/(

We add all the numbers together, and all the variables

(+k^2+1*4)/()+1

We multiply all the terms by the denominator


(+k^2+1*4)+1*()

We add all the numbers together, and all the variables

(+k^2+1*4)

We get rid of parentheses

k^2+1*4

We add all the numbers together, and all the variables

k^2+4

Back to the equation:

+(k^2+4)


We get rid of parentheses

k^2+4=0

a = 1; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·1·4
Δ = -16
Delta is less than zero, so there is no solution for the equation