What are the roots of x2 + 10x + 29 = 0?

a. 5 ± 2i

b. -5 ± 2i

c. 10 + 4i

d. -10 + 4i

B

Step-by-step explanation:

okay so x^2+10x+29=0

subtract 29 from both sides giving you:

x^2+10=-29

next, used (b/2)^2 to create a trinomial square, which will be 5 so you will have:

x^2-10x+(-5)^2=-29+(-5)^2

Then, simplify the equation:

x^2-10x+25=-4

Next, factor into (x-5)^2 which is:

(x-5)^2= -4

Finally, solve for X:

x=5 +/- 2i

the least common denominator is 40

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.