(2x^2-4x)=(x^2-3x-5)

Solution for (2x^2-4x)=(x^2-3x-5) equation:

(2x^2-4x)=(x^2-3x-5)

We move all terms to the left:

(2x^2-4x)-((x^2-3x-5))=0

We get rid of parentheses

2x^2-4x-((x^2-3x-5))=0


We calculate terms in parentheses: -((x^2-3x-5)), so:

(x^2-3x-5)

We get rid of parentheses

x^2-3x-5

Back to the equation:

-(x^2-3x-5)


We get rid of parentheses

2x^2-x^2-4x+3x+5=0

We add all the numbers together, and all the variables

x^2-1x+5=0

a = 1; b = -1; c = +5;
Δ = b2-4ac
Δ = -12-4·1·5
Δ = -19
Delta is less than zero, so there is no solution for the equation