Question

How do you find the roots, real and imaginary, of `y= 5x^2-6x +35` using the quadratic formula?

Answer 1

`x = 3/5 +- (sqrt(166)i)/5`

Explanation:

Use the quadratic formula `x = (-B +- sqrt(B^2 - 4AC))/(2A)` when the quadratic equation is in the form `Ax^2 + Bx +C = 0`

For `y = 5x^2 - 6x + 35 = 0, " "A = 5, B = -6, C = 35`

`x = (-B +- sqrt(B^2 - 4AC))/(2A) = (6 +- sqrt(36 - 4(5)(35)))/(2*5)`

`x = (6 +- sqrt(-664))/10 = 3/5 +- sqrt(-4*166)/10 = 3/5 +- (2 sqrt(166) i)/10`

`x = 3/5 +- (sqrt(166)i)/5`