Question

$2 {x}^{2} - 3x + 1$
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Answer 1

$2 {x}^{2} - 3x - 1 = 0 \\ d = {b}^{2} - 4ac \\ \: \: \: = { (- 3)}^{2} - 4(2)( - 1) \\ \: \: \: = 17$

now using the quadratic formula:

$= \frac{ - b \: (+ -) \: \sqrt{d} }{2a} \\ \\ = \frac{3 \: ( + - ) \sqrt{17} }{4}$

the two roots are=

$\frac{3 + \sqrt{17} }{4} \: \: \: and \: \frac{3 - \sqrt{17} }{4}$

PLZ MARK BRAINLIEST