Answer :

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

Here LHS = (2x – 1)(x - 3) = 2x^{2} –x - 6x + 3 = 2x^{2} - 7x + 3

RHS = (x + 5)(x - 1) = x^{2} + 5x – x - 5 = x^{2} + 4x - 5

Equating LHS and RHS

2x^{2} - 7x + 3 = x^{2} + 4x - 5

x^{2} – 11x + 8 = 0

which is of the form ax^{2} + bx + c = 0

∴ it is a quadratic equation.

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