Question

How do you solve `(x+3)(x+5) = -7`?

Answer 1

`x=-4+-isqrt(6)`

Explanation:

You first need to make one side `0`.

`[1]" "(x+3)(x+5)=-7`

Expand `(x+3)(x+5)`.

`[2]" "x^2+8x+15=-7`

Add `7` to both sides.

`[3]" "x^2+8x+15+7=-7+7`

`[4]" "x^2+8x+22=0`

Now that you have equated everything to `0`, you can solve for the roots. The roots of this quadratic equation are actually imaginary, so you should make use of the quadratic formula.

`[5]" "x=(-b+-sqrt(b^2-4ac))/(2a)`

`[6]" "x=(-(8)+-sqrt((8)^2-4(1)(22)))/(2(1))`

`[7]" "x=(-8+-sqrt(64-88))/2`

`[8]" "x=(-8+-sqrt(-24))/2`

`[9]" "x=(-8+-2isqrt(6))/2`

`[10]" "color(blue)(x=-4+-isqrt(6))`