How do you use the quadratic formula to solve `-x^2+1=-6x^2-x`?

So first off, we have to transform the equation to standard form.

`f(x)=-x^2+1=-6x^2-x`

To do this, we have to bring all the terms to one side and equate the equation to 0.

`f(x)=-x^2 + 6x^2+x+1=0`

Now we add like terms.

`f(x)=5x^2+x+1=0`

Once we have our equation in standard form, we use the quadratic formula: `x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}`.

We sub in the `a`, `b`, and `c` values. Not the variables - `x`.

`x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}`

`x=\frac{-(1)\pm\sqrt{(1)^2-4(5)(1)\ }}{2(5)}`

`x=\frac{-1\pm\sqrt{-19\ }}{10}`

I stopped here, because under the radical, we have a negative number. It's impossible to root a negative number - it becomes undefined.

The value under the radical indicates the number of zeros the equation has. A positive number greater than `0`, means there are two zeros. A value of `0` indicates there is only 1 zero. If there is a negative number under the radical, it means there are no zeros.

Therefore, the solution to this equation `f(x)=-x^2+1=-6x^2-x`, is that there are none. Without zeros, there are no solutions.

Hope this helps :)