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 add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

20x^2+17001

Back to the equation:

-(20x^2+17001)


We get rid of parentheses

-20x^2-17001=0

a = -20; b = 0; c = -17001;
Δ = b2-4ac
Δ = 02-4·(-20)·(-17001)
Δ = -1360080
Delta is less than zero, so there is no solution for the equation