Question

How do you find the zeros, real and imaginary, of `y=x^2+3x-4` using the quadratic formula?

Answer 1

The quatratic formula would be `(-b+-sqrt(b^2-4ac))/(2a)`

Explanation:

In this case `a=1,b=3,c=-4`
Using the formula we get:
`x_(1,2)=(-3+-sqrt(3^2-4*1*(-4)))/(2*1)`

`x_(1,2)=(-3+-sqrt(9+16))/2=(-3+-5)/2`

So: `x_1=-4and x_2=1`

Extra :
This could have been done easier by factoring:
`=(x-1)(x+4)`