57=(2x)^2/(.1-x)(.1-x)

Solution for 57=(2x)^2/(.1-x)(.1-x) equation:

57=(2x)^2/(.1-x)(.1-x)

We move all terms to the left:

57-((2x)^2/(.1-x)(.1-x))=0


Domain of the equation: (.1-x)(.1-x))!=0

We move all terms containing x to the left, all other terms to the right

-x)(.1-x)!=-.1

x∈R


We add all the numbers together, and all the variables

-(2x^2/(-1x+0.1)(-1x+0.1))+57=0

We multiply parentheses ..

-(2x^2/(+x^2-0.1x-0.1x+0.01))+57=0

We multiply all the terms by the denominator


-(2x^2+57*(+x^2-0.1x-0.1x+0.01))=0


We calculate terms in parentheses: -(2x^2+57*(+x^2-0.1x-0.1x+0.01)), so:

2x^2+57*(+x^2-0.1x-0.1x+0.01)

We multiply parentheses

2x^2+57x^2+0x+0x+0.57

We add all the numbers together, and all the variables

59x^2+2x+0.57

Back to the equation:

-(59x^2+2x+0.57)


We get rid of parentheses

-59x^2-2x-0.57=0

a = -59; b = -2; c = -0.57;
Δ = b2-4ac
Δ = -22-4·(-59)·(-0.57)
Δ = -130.52
Delta is less than zero, so there is no solution for the equation