Question

How do you find the roots, real and imaginary, of `y=-15^2 +40x -34` using the quadratic formula?

How do you find the roots, real and imaginary, of `y=-15x^2 +40x -34` using the quadratic formula?

Answer 1

Quadratic Formula is `x=(-b+-sqrt(b^2-4ac))/(2a)`

Explanation:

Where:
`a=-15`
`b=40`
`c=-34`

`:.x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-(40)+-sqrt((40)^2-4(-15)(-34)))/(2(-15))`
`x=(-40+-sqrt(-440))/(-30)`

And because one cannot square root any negative integers,
the answer obtained for `y=-15x^2+40x-34` would remain as:
`x=(-40+-sqrt(-440))/(-30)`

By presenting the roots:
`x=(-40+sqrt(-440))/(-30)` and `x=(-40-sqrt(-440))/(-30)`