How do you use the quadratic formula to find both solutions to the quadratic equation `x^2-3x=-10`?

`D = d^2 = b^2 - 4ac = 9 - 40 = -31 < 0` --> `d = +- isqrt31`

There are no real roots. There are 2 complex roots.

`x = 3/2 +-( isqrt31)/2`

I'm not sure the sign of `10` on the right hand side is correct, so let's deal with both cases:

Case 1 `x^2-3x=-10`

Add `10` to both sides to get `x^2-3x+10 = 0`

Let `f(x) = x^2-3x+10`.

This is of the form `ax^2+bx+c` with `a=1`, `b=-3` and `c=10`.

The discriminant `Delta` is given by the formula:

`Delta = b^2 - 4ac = (-3)^2-(4xx1xx10) = 9-40 = -31`

Since this is negative `f(x) = 0` has two distinct complex roots, given by the formula:

`x = (-b +-sqrt(Delta))/(2a) = (3+-i sqrt(31))/2`

Case 2 `x^2-3x=10`

Subtract `10` from both sides to get `x^2-3x-10 = 0`

Let `f(x) = x^2-3x-10`

This is of the form `ax^2+bx+c` with `a=1`, `b=-3` and `c=-10`

The discriminant `Delta` is given by the formula:

`Delta = b^2 - 4ac = (-3)^2-(4xx1xx-10) = 9+40 = 49 = 7^2`

This is positive and a perfect square, so the roots of `f(x) = 0` are distinct rational real numbers, given by the formula:

`x = (-b+-sqrt(Delta))/(2a) = (3+-7)/2`

That is `x = -4/2 = -2` and `x = 10/2 = 5`