Question

How do you solve `-5n ^ { 2} - 6n + 56= 0`?

Answer 1

`x1 = 14/5`
x2 = - 4

Explanation:

`f(n) = - y = - (5n^2 + 6n - 56) = 0`
Use the new Transforming Method to solve (Google Search)
Transformed equation
`y' = n^2 + 6n - 280 = 0`
Method: Find 2 real roots of y', then, divide them by a = 5.
Find the factor pair of (ac = 280), that has as sum (- b = -6).
Compose factor pairs of (-280) --> (5, -56)(10, -28)(14, -20).
This last sum is -6 = -b. Therefore, the 2 real roots of y' are:
14 and - 20.
Back to original y, the 2 real roots are:
`x1 = 14/(a) = 14/5`, and `x2 = -20/a = -20/5 = - 4`
Note. It is not necessary to factor by grouping and solving the 2 binomials.