solve by useing the quadratic formula x^2 + 5x=-8?

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x=-5+-sqrt(25-32)/2=-5+-isqrt7 all over 2

To use the quadratic formula, one first writes the equation in the standard format that looks like this:

a x^2 + b x + c = 0

In your case, this would be
x^2 + 5x + 8 = 0

a = 1
b = 5
c = 8

The quadratic formula is

x = [ -b +/- SQRT( b^2 - 4ac ) ] / 2a

The portion under the root: (b^2 - 4ac)
is called the determinant, because it "determines" the type of answer you get.

If the determinant is 0, there is only one root (later, you will learn that it is called a root of multiplicity two).
If the determinant is positive, then there are two real answers.
If the determinant is negative, there is no real value for x (there are "complex" values that satisfy the equation).

So, back to the formula:
x = [ -b +/- SQRT( b^2 - 4ac ) ] / 2a

with a =1 b=5 c=8
(b^2 - 4ac) = 25 - 32 = -7
negative = no "real" answer, only complex values.

x = [5 +/- SQRT(-7)]/2

x = (1/2)*[5 + SQRT(-7)]
and
x = (1/2)*[5 - SQRT(-7)]

(If you are working with real numbers, then the SQRT(-7) does not exist)

x^2 + 5x + 8 = 0

(-5 pm sqrt((5^2) - (4 * 8))/ 2 (pm is plus or minus)

-5/2 pm (sqrt(-7)/2) <<<<< no real solutions!

x^2 + 5x = -8
x^2 + 5x + 8 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 5
c = 8

x = [-5 ±√(25 - 32)]/2
x = [-5 ±√-7]/2 (imaginary number)
(no real roots)