Question

find the root.
$x = (- b + - \sqrt{ {b}^{2} } - 4ac \\ \: ) \div 2a$
​

Answer 1

$x = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - ( - 60) \frac{ + }{} \sqrt{ {60}^{2} - 4 \times 36 \times 23 } }{2 \times 36} \\ x = \frac{60 \frac{ + }{} \sqrt{3600 - 3312} }{120} \\ x = \frac{60 \frac{ + }{} \sqrt{288} }{120}$

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Answer 2

Answer:

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Step-by-step explanation:

$36 {x}^{2} + 23 = 60x \\ 36 {x}^{2} - 60x + 23 = 0 \\ a {x}^{2} + bx + c = 0 \\ a = 36 \\ b = - 60 \\ c = 23 \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \: \: \: = \frac{ - ( - 60) + - \sqrt{ { - 60}^{2} - 4 \times 36 \times 23 } }{2 \times 36} \\ \: \: \: = \frac{60 + - \sqrt{3600 - 3312} }{72} \\ \: \: \: = \frac{60 + - 12 \sqrt{2} }{72} \\ = \frac{12(5 + - \sqrt{2} )}{72} \\ = \frac{5 + - \sqrt{2} }{6} \\ then \\ \alpha = \frac{5 + \sqrt{2} }{6} ansssssss \\ \beta = \frac{5 - \sqrt{2} }{6} ansssssss$