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Katyanochek1 [597]
3 years ago
13

Find all complex solutions of 2x^2-x+8=0

Mathematics
1 answer:
kumpel [21]3 years ago
7 0

Answer:

x = 1 − √63i / 4 , x = 1 + √63i / 4

Step-by-step explanation:

Let's use the formula −b±√b2−(4ac) / 2⋅a

=  ax^2+bx+c=0

= 2x^2-x+8=0

Let's substitute 2 for a, −1 for b, and 8 for c in the formula of quadratic equation.

x = −1±√−12−(4⋅2⋅8) / 2⋅2

x = 1±√1−64 / 4

x = 1±√−63 / 4

x = 1±√63⋅−1 / 4

x = 1±√63√−1 / 4

x = {1−√63i / 4 , 1±√63i / 4}

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