Question

How do you use the quadratic formula to find both solutions to the quadratic equation `x^2+3x-4=0`?

Answer 1

By replacing the variable coefficients in the general quadratic formual with the coefficients for the terms in the given equation we can get `x=1` or `x=-4`

Explanation:

For an equation in the form:
`color(white)("XXXX")` `ax^2+bx+c=0`
the quadratic formula gives the solutions as
`color(white)("XXXX")` `x= (-b+-sqrt(b^2-4ac))/(2a)`

`x^2+3x-4 = 0` is in this form with
`color(white)("XXXX")` `a=1`
`color(white)("XXXX")` `b=3` and
`color(white)("XXXX")` `c=(-4)`

Replacing `a, b, and c` in the quadratic formula, we get
`color(white)("XXXX")` `x=(-3+-sqrt(3^2-4(1)(-4)))/(2(1))`

`color(white)("XXXX")` `x= (-3+-sqrt(9+16))/2`

`color(white)("XXXX")` `x=(-3+-5)/2`

`color(white)("XXXX")` `x = 2/2` or `x= (-8)/2`

`x=1` or `x=-4`

Answer 2

`x_1=-4`
`x_2=1`

Explanation:

Having the equation `x^2+3x−4=0`, we find the `Delta_x`:
`Delta_x=3^2-4*1*(-4)=9+16=25`

Now we find `x_1` and `x_2` with the quadratic formula .

`x_"1,2"=(-3+-sqrt(Delta_x))/(2*1)=(-3+-5)/2`.

The two solutions are:

`x_1=(-3-5)/2=-4`

`x_2=(-3+5)/2=1`.