2math questions quadratic equations

The third one.

If we say that a quadratic eqtn takes the form of ax^2 + bx + c = 0, then in this eqtn a=1, b=-16 and c=73. Since the formula for finding the roots of quadratic eqtns is:

(-b+-sqrt(b^2-4ac))/2a, then:

(16 +- sqrt(16^2-4*73*1))/2

(16 +- sqrt(256-292))/2

(16 +- sqrt(-36))/2

Now, we know that the imaginary number “i” is equal to the square root of minus one. If we break down sqrt(-36) to sqrt(-1)*sqrt(36), then we get “6i” because the sqrt of 36 is 6 and the sqrt of -1 is i.

Now we have:

(16 +- 6i)/2 — if we divide this expression by two, we get 8 +- 3i.

5\16 x 2\5 = 10/80simplify to 1/80