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(2x-1)/3=5x^2
We move all terms to the left:
(2x-1)/3-(5x^2)=0
determiningTheFunctionDomain -5x^2+(2x-1)/3=0
We multiply all the terms by the denominator
-5x^2*3+(2x-1)=0
Wy multiply elements
-15x^2+(2x-1)=0
We get rid of parentheses
-15x^2+2x-1=0
a = -15; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·(-15)·(-1)
Δ = -56
Delta is less than zero, so there is no solution for the equation
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