Question

simply it please......​

Answer 1

Step-by-step explanation:

$({x}^{2} - x) - \frac{1}{2} (x - 3 + 3 {x}^{2} ) = 0\\ \\ = > 2 {x}^{2} - 2x - x + 3 - 3{x}^{2} = 0 \\ \\ \\ = > - {x}^{2} - 3x + 3 = 0 \\ \\ = > {x}^{2} + 3x - 3 = 0 \\ \\ d = \sqrt{ {b}^{2} - 4ac} \\ \\hear \\ a = 1 \\ b = 3 \\ c = ( - 3) \: \\ so.. \\ d = \sqrt{ {3}^{2} - 4 \times 1 \times ( - 3)} \\ \\ d = \sqrt{9 + 12} = \sqrt{21} \\ \\ \\ \\ then \: you \: know.. \\ \: x = \frac{ - b + - \sqrt{d} }{2a} \\ \\ \\ \\ x = \frac{ - 3 + - \sqrt{21} }{2 \times 1} \\ \\ x= > \frac{ - 3 + \sqrt{21} }{2} \\ and \\ \\ x = > \frac{ - 3 - \sqrt{21} }{2}$

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Answer 2

Answer:

answer is correct most correct