(2x^2+3x+5=7x-1)

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Solution for (2x^2+3x+5=7x-1) equation: 



(2x^2+3x+5=7x-1)

We move all terms to the left:

(2x^2+3x+5-(7x-1))=0



We calculate terms in parentheses: +(2x^2+3x+5-(7x-1)), so:

2x^2+3x+5-(7x-1)

determiningTheFunctionDomain
2x^2+3x-(7x-1)+5

We get rid of parentheses

2x^2+3x-7x+1+5

We add all the numbers together, and all the variables

2x^2-4x+6

Back to the equation:

+(2x^2-4x+6)



We get rid of parentheses

2x^2-4x+6=0

a = 2; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·2·6
Δ = -32
Delta is less than zero, so there is no solution for the equation

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