(x-3)2+(x+1)2+(3x-5)=2x2

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Solution for (x-3)2+(x+1)2+(3x-5)=2x2 equation: 



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

We move all terms to the left:

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

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

We multiply parentheses

-2x^2+2x+2x+(3x-5)-6+2=0

We get rid of parentheses

-2x^2+2x+2x+3x-5-6+2=0

We add all the numbers together, and all the variables

-2x^2+7x-9=0

a = -2; b = 7; c = -9;
Δ = b2-4ac
Δ = 72-4·(-2)·(-9)
Δ = -23
Delta is less than zero, so there is no solution for the equation

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