Question

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

Answer 1

$(3x - 1)2 - 5(x - 2) - (2x - 3)2 - (x + 5)(x - 3) = 0 \\ 6x - 2 - 5x + 10 - 4x + 6 - ( {x}^{2} + 2x - 15) = 0 \\ - 3x + 14 - {x}^{2} - 2 x + 15 = 0 \\ - {x}^{2} - 5x + 29 = 0 \\ 0 = {x}^{2} + 5x - 29 \\ y \: esto \: lo \: puedes \: resolver \: por \: la \: formula \: general \\ \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ remplazando \: los \: terminos \\ = \frac{ - 5 + \sqrt{ {5}^{2} - 4(1)( - 29)} }{2(1)} \\ = \frac{ - 5 + \sqrt{141} }{2} \\ = \frac{ - 5 - \sqrt{141} }{2}$
saludos

Answer 2

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

$9x^{2} - 6x + 1 - 5x + 10 - (4 x^{2} -12x + 9) - (x^{2} +2x - 15) = 0$

$9x^{2} - 6x + 1 - 5x + 10 - 4 x^{2} + 12x - 9 - x^{2} -2x + 15 = 0$

$4x^{2} + 12x - 13x + 26 - 9 = 0$

$4x^{2} - 1x + 17 = 0$