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

1 Answer
Oct 27, 2015

Answer:

#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))#