68=(1/5)(20x+50)+2

Solution for 68=(1/5)(20x+50)+2 equation:

68=(1/5)(20x+50)+2

We move all terms to the left:

68-((1/5)(20x+50)+2)=0


Domain of the equation: 5)(20x+50)+2)!=0

x∈R


We add all the numbers together, and all the variables

-((+1/5)(20x+50)+2)+68=0

We multiply parentheses ..

-((+20x^2+1/5*50)+2)+68=0

We multiply all the terms by the denominator


-((+20x^2+1+68*5*50)+2)=0


We calculate terms in parentheses: -((+20x^2+1+68*5*50)+2), so:

(+20x^2+1+68*5*50)+2

We get rid of parentheses

20x^2+1+2+68*5*50

We add all the numbers together, and all the variables

20x^2+17003

Back to the equation:

-(20x^2+17003)


We get rid of parentheses

-20x^2-17003=0

a = -20; b = 0; c = -17003;
Δ = b2-4ac
Δ = 02-4·(-20)·(-17003)
Δ = -1360240
Delta is less than zero, so there is no solution for the equation