(1/5)(x+2)=-3

Solution for (1/5)(x+2)=-3 equation:

(1/5)(x+2)=-3

We move all terms to the left:

(1/5)(x+2)-(-3)=0


Domain of the equation: 5)(x+2)!=0

x∈R


We add all the numbers together, and all the variables

(+1/5)(x+2)-(-3)=0

We add all the numbers together, and all the variables

(+1/5)(x+2)+3=0

We multiply parentheses ..

(+x^2+1/5*2)+3=0

We multiply all the terms by the denominator


(+x^2+1+3*5*2)=0

We get rid of parentheses

x^2+1+3*5*2=0

We add all the numbers together, and all the variables

x^2+31=0

a = 1; b = 0; c = +31;
Δ = b2-4ac
Δ = 02-4·1·31
Δ = -124
Delta is less than zero, so there is no solution for the equation